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Tautological rings for high-dimensional manifolds

Published online by Cambridge University Press:  13 March 2017

Søren Galatius
Department of Mathematics, Stanford University, Stanford CA, 94305, USA email
Ilya Grigoriev
Department of Mathematics, University of Chicago, 5734 S. University Avenue, Chicago, IL 60637, USA email
Oscar Randal-Williams
Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK email
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We study tautological rings for high-dimensional manifolds, that is, for each smooth manifold $M$ the ring $R^{\ast }(M)$ of those characteristic classes of smooth fibre bundles with fibre $M$ which is generated by generalised Miller–Morita–Mumford classes. We completely describe these rings modulo nilpotent elements, when $M$ is a connected sum of copies of $S^{n}\times S^{n}$ for $n$ odd.

Research Article
© The Authors 2017 

1 Introduction

Let $E^{k+d}$ and $B^{k}$ be connected compact smooth oriented manifolds, and let $\unicode[STIX]{x1D70B}:E\rightarrow B$ be a smooth fibre bundle (i.e. a proper submersion) with fibre $M^{d}$ . Then there is a $d$ -dimensional oriented vector bundle $T_{\unicode[STIX]{x1D70B}}=\text{Ker}(D\unicode[STIX]{x1D70B})$ over $E$ . If $c\in H^{\ast }(B\text{SO}(d);\mathbb{Q})$ is a rational characteristic class of such vector bundles, we define the associated generalised Miller–Morita–Mumford class to be the fibre integral

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{c}(\unicode[STIX]{x1D70B})=\int _{\unicode[STIX]{x1D70B}}c(T_{\unicode[STIX]{x1D70B}})\in H^{\ast -d}(B;\mathbb{Q}).\end{eqnarray}$$

For a basis ${\mathcal{B}}\subset H^{\ast }(B\text{SO}(d);\mathbb{Q})$ this defines a ring homomorphism

$$\begin{eqnarray}\displaystyle \mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}] & \longrightarrow & \displaystyle H^{\ast }(B;\mathbb{Q})\nonumber\\ \displaystyle \unicode[STIX]{x1D705}_{c} & \longmapsto & \displaystyle \unicode[STIX]{x1D705}_{c}(\unicode[STIX]{x1D70B}).\nonumber\end{eqnarray}$$

We let $I_{M}\subset \mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]$ be the ideal consisting of those polynomials in the classes $\unicode[STIX]{x1D705}_{c}$ which vanish on every such smooth fibre bundle with fibre $M$ , and define the tautological ring of $M$ to be the associated quotient ring

$$\begin{eqnarray}R^{\ast }(M)=\mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]/I_{M}.\end{eqnarray}$$

The name ‘tautological ring’ is borrowed from the $d=2$ case, where it usually refers to the subring of the cohomology ring (or Chow ring) of moduli spaces of Riemann surfaces which is generated by certain tautological classes $\unicode[STIX]{x1D705}_{i}$ (which in our notation correspond to $\unicode[STIX]{x1D705}_{e^{i+1}}$ ). Our definition coincides with the usual one in this case. There is a large literature on these rings; see [Reference MumfordMum83, Reference LooijengaLoo95, Reference FaberFab99, Reference MoritaMor03].

We shall be interested in manifolds of even dimension $d=2n$ , in which case $H^{\ast }(B\text{SO}(2n);\mathbb{Q})=\mathbb{Q}[p_{1},p_{2},\ldots ,p_{n-1},e]$ and we take ${\mathcal{B}}$ to be the basis of monomials in these classes. Writing $W_{g}$ for the connected sum of $g$ copies of $S^{n}\times S^{n}$ , our goal in this paper is to study the structure of $R^{\ast }(W_{g})$ modulo the nilradical $\sqrt{0}$ , i.e. the ideal of nilpotent elements, and our main result can be stated as follows.

Theorem 1.1. Let $n$ be odd. Then:

  1. (i) $R^{\ast }(W_{0})/\sqrt{0}=\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}]$ ;

  2. (ii) $R^{\ast }(W_{1})/\sqrt{0}=\mathbb{Q}$ ;

  3. (iii) $R^{\ast }(W_{g})/\sqrt{0}=\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n-1}}]$ for $g>1$ .

In fact, (i) already holds before taking the quotient by the nilradical, and it holds for $n$ both even and odd (see § 5.3).

As the Krull dimension of a ring is unchanged upon taking the quotient by the nilradical, we may also conclude that $R^{\ast }(W_{0})$ has Krull dimension $n$ , $R^{\ast }(W_{1})$ has Krull dimension 0, and $R^{\ast }(W_{g})$ has Krull dimension $n-1$ as long as $g>1$ .

We shall also study two closely related tautological rings. Firstly, we may consider fibre bundles $\unicode[STIX]{x1D70B}:E^{k+2n}\rightarrow B^{k}$ with fibre $M^{2n}$ equipped with a section $s:B\rightarrow E$ . To such a fibre bundle and a $c\in {\mathcal{B}}$ we may also associate the characteristic class $c(\unicode[STIX]{x1D70B},s)=s^{\ast }(c(T_{\unicode[STIX]{x1D70B}}))$ , so in this case a ring homomorphism is defined,

$$\begin{eqnarray}\displaystyle \mathbb{Q}[c,\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}] & \longrightarrow & \displaystyle H^{\ast }(B;\mathbb{Q})\nonumber\\ \displaystyle c & \longmapsto & \displaystyle c(\unicode[STIX]{x1D70B},s)\nonumber\\ \displaystyle \unicode[STIX]{x1D705}_{c} & \longmapsto & \displaystyle \unicode[STIX]{x1D705}_{c}(\unicode[STIX]{x1D70B}),\nonumber\end{eqnarray}$$

and we let $I_{(M,\star )}\subset \mathbb{Q}[c,\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]$ denote the ideal of those polynomials in the $c$ and $\unicode[STIX]{x1D705}_{c}$ which vanish on all such bundles; we write $R^{\ast }(M,\star )$ for the associated quotient ring.

Theorem 1.2. Let $n$ be odd.

  1. (i) For each $c\in {\mathcal{B}}$ , $(2-2g)\cdot c=\unicode[STIX]{x1D705}_{ec}\in R^{\ast }(W_{g},\star )/\sqrt{0}$ .

  2. (ii) For $g\neq 1$ , the map

    $$\begin{eqnarray}R^{\ast }(W_{g})/\sqrt{0}\longrightarrow R^{\ast }(W_{g},\star )/\sqrt{0}\end{eqnarray}$$
    is an isomorphism.
  3. (iii) The class $e\in R^{\ast }(W_{1},\star )$ is nilpotent, and the map

    $$\begin{eqnarray}\mathbb{Q}[p_{1},p_{2},\ldots ,p_{n-1}]\longrightarrow R^{\ast }(W_{1},\star )/\sqrt{0}\end{eqnarray}$$
    is an isomorphism.

Secondly, we may consider fibre bundles $\unicode[STIX]{x1D70B}:E^{k+2n}\rightarrow B^{k}$ with fibre $M^{2n}$ equipped with an embedding $S:D^{2n}\times B\rightarrow E$ over $B$ (or, equivalently, with a normally framed section). The restriction $s=S|_{\{0\}\times B}$ defines a section, but the fact that it is framed means that $c(\unicode[STIX]{x1D70B},s)=0$ . Hence in this case we denote by $I_{(M,D^{2n})}\subset \mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]$ the ideal of polynomials in the $\unicode[STIX]{x1D705}_{c}$ which vanish on all such bundles, and write $R^{\ast }(M,D^{2n})$ for the associated quotient ring. The following is immediate from Theorems 1.1 and 1.2(i).

Corollary 1.3. If $n$ is odd, then $R^{\ast }(W_{g},D^{2n})/\sqrt{0}=\mathbb{Q}$ for all $g$ .

The second author has shown (see [Reference GrigorievGri13, Theorem 1.1]) that for $n$ odd the ring $R^{\ast }(W_{g})$ is finitely generated as a $\mathbb{Q}$ -algebra, and hence $R^{\ast }(W_{g},D^{2n})$ is too. (In fact, [Reference GrigorievGri13, Theorem 1.1] only shows this for $g>1$ , but in Proposition 5.7 we will show that $R^{\ast }(W_{g},\star )$ , and hence $R^{\ast }(W_{g},D^{2n})$ , is finitely generated for all $g$ .) Corollary 1.3 therefore implies the following result.

Corollary 1.4. If $n$ is odd, then $R^{\ast }(W_{g},D^{2n})$ is a finite-dimensional rational vector space for all  $g$ .

It would be interesting to have a computation of the ring $R^{\ast }(W_{g},D^{2n})$ for some $g>1$ . In proving these theorems, we obtain results about the vanishing of certain elements in $R^{\ast }(M)/\sqrt{0}$ for any manifold $M$ (in Theorem 2.1) and about the algebraic independence of certain elements in $R^{\ast }(W_{g})/\sqrt{0}$ for $n$ even (in Theorem 4.1). We cannot obtain results as conclusive as Theorem 1.1 for $n$ even, as our argument relies on [Reference GrigorievGri13], which does not apply in this case.

2 Nilpotence of the Hirzebruch ${\mathcal{L}}$ -classes

In this section, we prove that certain generalised Miller–Morita–Mumford classes are nilpotent in $R^{\ast }(M)$ for any smooth even-dimensional manifold $M^{2n}$ . We will later apply this together with results from [Reference GrigorievGri13] to the manifolds $W_{g}$ in order to obtain an upper bound on $R^{\ast }(W_{g})/\sqrt{0}$ .

2.1 (Modified) Hirzebruch ${\mathcal{L}}$ -classes

We first define certain cohomology classes $\tilde{{\mathcal{L}}}_{i}\in H^{4i}(B\text{SO}(2n);\mathbb{Q})$ . Consider the graded ring $\mathbb{Q}[x_{1},\ldots ,x_{n}]$ in which each variable $x_{1},\ldots ,x_{n}$ has degree 2. We define homogenous symmetric polynomials $\tilde{{\mathcal{L}}}_{i}$ by the expression

$$\begin{eqnarray}\tilde{{\mathcal{L}}}=2^{n}+\tilde{{\mathcal{L}}}_{1}+\tilde{{\mathcal{L}}}_{2}+\cdots =\mathop{\prod }_{i=1}^{n}\frac{x_{i}}{\operatorname{tanh}x_{i}/2}.\end{eqnarray}$$

(The function $x/(\operatorname{tanh}x/2)$ is even, so the above expresses $\tilde{{\mathcal{L}}}_{i}$ as a symmetric polynomial in $x_{1}^{2},x_{2}^{2},\ldots ,x_{n}^{2}$ .) The subring of symmetric polynomials is generated by the elementary symmetric polynomials $\unicode[STIX]{x1D70E}_{i,n}(x_{1}^{2},\ldots ,x_{n}^{2})$ , so we may express $\tilde{{\mathcal{L}}}_{i}$ as a polynomial $\tilde{{\mathcal{L}}}_{i}(\unicode[STIX]{x1D70E}_{1,n},\ldots ,\unicode[STIX]{x1D70E}_{n,n})$ , and we write

$$\begin{eqnarray}\tilde{{\mathcal{L}}}_{i}=\tilde{{\mathcal{L}}}_{i}(p_{1},p_{2},\ldots ,p_{n})\in H^{4i}(B\text{SO}(2n);\mathbb{Q})\end{eqnarray}$$

for this polynomial evaluated at the Pontrjagin classes. Note that these differ from the usual Hirzebruch $L$ -classes, ${\mathcal{L}}_{i}$ , by a factor: $\tilde{{\mathcal{L}}}_{i}=2^{n-2i}{\mathcal{L}}_{i}$ . These classes may be written as

$$\begin{eqnarray}\tilde{{\mathcal{L}}}_{i}=2^{n}(2^{2i-1}-1)\frac{B_{i}}{(2i)!}\cdot p_{i}+(\text{polynomial in lower Pontrjagin classes})\end{eqnarray}$$

where $B_{i}$ is the $i$ th Bernoulli number; cf. [Reference Milnor and StasheffMS74, Problem 19-C] (this uses the convention of [Reference Milnor and StasheffMS74, Appendix B], in which $B_{i}=(-1)^{i}\cdot 2i\cdot \unicode[STIX]{x1D701}(1-2i)$ ). In particular, these leading coefficients are never zero, so $\tilde{{\mathcal{L}}}_{1},\tilde{{\mathcal{L}}}_{2},\ldots ,\tilde{{\mathcal{L}}}_{n}$ generate the same subring of $H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ as $p_{1},p_{2},\ldots ,p_{n}$ .

2.2 Nilpotence due to the Hirzebruch signature theorem

For any manifold $M^{2n}$ , the characteristic classes $\tilde{{\mathcal{L}}}_{i}\in H^{4i}(B\text{SO}(2n);\mathbb{Q})$ give rise to the corresponding generalised Miller–Morita–Mumford classes $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}\in R^{\ast }(M)$ . Our present goal is to prove the following.

Theorem 2.1. The classes $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}\in R^{\ast }(M)$ are nilpotent for all natural numbers $i\geqslant 1$ such that $4i-2n\neq 0$ (all natural numbers if $n$ is odd).

The main ingredient in the proof is a parametrised version of the Hirzebruch signature theorem, due to Atiyah [Reference AtiyahAti69], which relates the classes $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}$ to classes arising from the cohomology of an arithmetic group.

Let $M^{2n}$ be an oriented manifold, and let $H=H^{n}(M;\mathbb{Z})/\text{torsion}$ . It is a consequence of Poincaré duality that the $(-1)^{n}$ -symmetric pairing

$$\begin{eqnarray}\unicode[STIX]{x1D706}:H\otimes _{\mathbb{Z}}H\stackrel{\cup }{\longrightarrow }H^{2n}(M;\mathbb{Z})/\text{torsion}=H^{2n}(M;\mathbb{Z})\stackrel{{\sim}}{\longrightarrow }\mathbb{Z}\end{eqnarray}$$

is non-degenerate. Any homotopy equivalence of $M$ preserving its orientation therefore induces an isometry of the $(-1)^{n}$ -symmetric form $(H,\unicode[STIX]{x1D706})$ . In particular, if $\unicode[STIX]{x1D70B}:E\rightarrow B$ is a smooth oriented fibre bundle with fibre $M$ , there is a homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}:\unicode[STIX]{x1D70B}_{1}(B)\longrightarrow \text{Aut}(H,\unicode[STIX]{x1D706})\end{eqnarray}$$

from the fundamental group of $B$ to the group of isometries of $(H,\unicode[STIX]{x1D706})$ , inducing a map $\unicode[STIX]{x1D719}:B\rightarrow B\text{Aut}(H,\unicode[STIX]{x1D706})$ .

We may further consider the induced $(-1)^{n}$ -symmetric form $\unicode[STIX]{x1D706}_{\mathbb{R}}$ on $H_{\mathbb{R}}=H\otimes _{\mathbb{Z}}\mathbb{R}$ . There are now two cases, depending on the parity of $n$ . If $n$ is odd, then $(H_{\mathbb{R}},\unicode[STIX]{x1D706}_{\mathbb{R}})$ is a non-degenerate skew-symmetric form over $\mathbb{R}$ , and so is determined by its rank. This identifies $\text{Aut}(H_{\mathbb{R}},\unicode[STIX]{x1D706}_{\mathbb{R}})$ with $\text{Sp}_{2g}(\mathbb{R})$ , where $H_{\mathbb{R}}$ has dimension $2g$ . If $n$ is even, then $(H_{\mathbb{R}},\unicode[STIX]{x1D706}_{\mathbb{R}})$ is a non-degenerate symmetric form over $\mathbb{R}$ , and so is determined by its rank and signature. This identifies $\text{Aut}(H_{\mathbb{R}},\unicode[STIX]{x1D706}_{\mathbb{R}})$ with $\text{O}_{p,q}(\mathbb{R})$ , where $H_{\mathbb{R}}$ has dimension $p+q$ and signature $p-q$ . We thus obtain a composition

$$\begin{eqnarray}B\overset{\unicode[STIX]{x1D719}}{\longrightarrow }B\text{Aut}(H,\unicode[STIX]{x1D706})\longrightarrow B\text{Aut}(H_{\mathbb{R}},\unicode[STIX]{x1D706}_{\mathbb{R}})=\left\{\begin{array}{@{}l@{}}B\text{Sp}_{2g}(\mathbb{R}),\quad \\ B\text{O}_{p,q}(\mathbb{R}).\quad \end{array}\right.\end{eqnarray}$$

Now Atiyah’s work [Reference AtiyahAti69, §4] implies that the classes $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}$ are pulled back via this composition, so in particular they are in the image of $\unicode[STIX]{x1D719}^{\ast }:H^{\ast }(B\text{Aut}(H,\unicode[STIX]{x1D706});\mathbb{Q})\rightarrow H^{\ast }(B;\mathbb{Q})$ . Theorem 2.1 is then an immediate consequence of the fact that all positive-degree elements of $H^{\ast }(B\text{Aut}(H,\unicode[STIX]{x1D706});\mathbb{Q})$ are nilpotent; this is because $\text{Aut}(H,\unicode[STIX]{x1D706})$ is an arithmetic group, and so has finite virtual cohomological dimension and hence in particular finite $\mathbb{Q}$ -cohomological dimension.

3 Relations modulo nilpotence

The main result (Theorem 2.7) of [Reference GrigorievGri13] implies that when $n$ is odd, for any oriented manifold bundle $E\rightarrow B$ with fibre $W_{g}$ and even-dimensional classes $a,b\in H^{\ast }(E;\mathbb{Q})$ such that $\unicode[STIX]{x1D70B}_{!}(a)=0$ and $\unicode[STIX]{x1D70B}_{!}(b)=0$ , the class $\unicode[STIX]{x1D70B}_{!}(ab)$ is nilpotent. We shall use the following stronger statement, which will be deduced quite formally from [Reference GrigorievGri13, Theorem 2.7].

Theorem 3.1. Let $\unicode[STIX]{x1D70B}:E\rightarrow B$ be a fibration with homotopy fibre homotopy-equivalent to $W_{g}=g(S^{n}\times S^{n})$ for $n$ odd, such that the action of $\unicode[STIX]{x1D70B}_{1}(B,b)$ on $H^{2n}(\unicode[STIX]{x1D70B}^{-1}(b);\mathbb{Q})$ is trivial. For classes $a,b\in H^{\ast }(E;\mathbb{Q})$ such that $A=|\unicode[STIX]{x1D70B}_{!}(a)|$ and $B=|\unicode[STIX]{x1D70B}_{!}(b)|$ are even, if $\unicode[STIX]{x1D70B}_{!}(a)^{k}=0$ and $\unicode[STIX]{x1D70B}_{!}(b)^{l}=0$ , then $\unicode[STIX]{x1D70B}_{!}(ab)^{(2g+1)(Ak+Bl)}=0$ .

If $g\neq 1$ , this is a consequence of [Reference GrigorievGri13, Lemma 5.7]. The proof we present here is independent of genus.

Proof. First, suppose that both $A$ and $B$ are at least $2$ . Let $F$ be defined by the homotopy fibre sequence

$$\begin{eqnarray}F\longrightarrow K(\mathbb{Q},A)\times K(\mathbb{Q},B)\overset{\unicode[STIX]{x1D704}_{A}^{k}\times \unicode[STIX]{x1D704}_{B}^{l}}{\longrightarrow }K(\mathbb{Q},kA)\times K(\mathbb{Q},lB),\end{eqnarray}$$

where the map between the Eilenberg–MacLane spaces is given by the product of the appropriate powers of the fundamental cohomology classes. If $A$ and $B$ are even, it is an elementary calculation to see that

$$\begin{eqnarray}H^{\ast }(F;\mathbb{Q})=\mathbb{Q}[\unicode[STIX]{x1D704}_{A},\unicode[STIX]{x1D704}_{B}]/(\unicode[STIX]{x1D704}_{A}^{k},\unicode[STIX]{x1D704}_{B}^{l})\end{eqnarray}$$

and that $F$ is also simply connected. Thus $F$ has $\mathbb{Q}$ -cohomological dimension $Ak+Bl-1$ .

As $\unicode[STIX]{x1D70B}_{!}(a)^{k}=0$ and $\unicode[STIX]{x1D70B}_{!}(b)^{l}=0$ , the map

$$\begin{eqnarray}(\unicode[STIX]{x1D70B}_{!}(a),\unicode[STIX]{x1D70B}_{!}(b)):B\longrightarrow K(\mathbb{Q},A)\times K(\mathbb{Q},B)\end{eqnarray}$$

representing $(\unicode[STIX]{x1D70B}_{!}(a),\unicode[STIX]{x1D70B}_{!}(b))$ lifts to a map $f:B\rightarrow F$ , and we let $B^{\prime }$ denote the homotopy fibre of $f$ and $i:B^{\prime }\rightarrow B$ the evaluation map. The class $i^{\ast }(\unicode[STIX]{x1D70B}_{!}(ab))$ is the same as $\unicode[STIX]{x1D70B}_{!}^{\prime }(a^{\prime }b^{\prime })$ , where $\unicode[STIX]{x1D70B}^{\prime }:E^{\prime }\rightarrow B^{\prime }$ is the pullback of $\unicode[STIX]{x1D70B}$ along $i$ , and $a^{\prime }$ and $b^{\prime }$ are the pullbacks of $a$ and $b$ along the map $E^{\prime }\rightarrow E$ covering $i$ . The classes $\unicode[STIX]{x1D70B}_{!}^{\prime }(a^{\prime })$ and $\unicode[STIX]{x1D70B}_{!}^{\prime }(b^{\prime })$ are trivial, so by [Reference GrigorievGri13, Theorem 2.7] we have $\unicode[STIX]{x1D70B}_{!}^{\prime }(a^{\prime }b^{\prime })^{2g+1}=0$ . Thus $\unicode[STIX]{x1D70B}_{!}(ab)^{2g+1}$ lies in the kernel of $i^{\ast }$ and so has positive Serre filtration with respect to the map $f$ . Hence $\unicode[STIX]{x1D70B}_{!}(ab)^{(2g+1)(Ak+Bl)}$ has Serre filtration at least $Ak+Bl$ , which is beyond the $\mathbb{Q}$ -cohomological dimension of $F$ , so it is zero as required.

Finally, if $A=0$ and $B=0$ , then the situation becomes that of [Reference GrigorievGri13, Theorem 2.7]; this leaves the case of $A=0$ and $B\geqslant 2$ . In this case we must have $\unicode[STIX]{x1D70B}_{!}(a)=0$ , and the above argument can be followed using the homotopy fibre of $\unicode[STIX]{x1D704}_{B}^{l}:K(\mathbb{Q},B)\rightarrow K(\mathbb{Q},Bl)$ in place of  $F$ .◻

Let us also recall some further results from [Reference GrigorievGri13], as well as a convenient relation between $R^{\ast }(W_{g})$ and $R^{\ast }(W_{g},\star )$ . We shall often write $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}(W_{g})=2-2g$ for the Euler characteristic of  $W_{g}$ .

Lemma 3.2. Let $n$ be odd.

  1. (i) For any $c\in {\mathcal{B}}$ , we have

    (1) $$\begin{eqnarray}\unicode[STIX]{x1D712}^{2}c-\unicode[STIX]{x1D712}\unicode[STIX]{x1D705}_{ec}-\unicode[STIX]{x1D712}e\unicode[STIX]{x1D705}_{c}+\unicode[STIX]{x1D705}_{e^{2}}\unicode[STIX]{x1D705}_{c}=0\in R^{\ast }(M_{g},\star )/\sqrt{0}.\end{eqnarray}$$
  2. (ii) We have

    (2) $$\begin{eqnarray}(\unicode[STIX]{x1D712}-2)\unicode[STIX]{x1D712}e+\unicode[STIX]{x1D705}_{e^{2}}=0\in R^{\ast }(M_{g},\star )/\sqrt{0}.\end{eqnarray}$$
  3. (iii) If $g\neq 1$ then the map

    $$\begin{eqnarray}R^{\ast }(W_{g})\longrightarrow R^{\ast }(W_{g},\star )\end{eqnarray}$$
    is injective.

Proof. Statement (i) is [Reference GrigorievGri13, Example 5.19] but taking $a=\unicode[STIX]{x1D712}\unicode[STIX]{x1D708}_{(1\ast )}-e_{(\ast )}$ and $b=\unicode[STIX]{x1D712}p_{(\ast )}-e_{(\ast )}\unicode[STIX]{x1D705}_{p}$ to avoid dividing by $\unicode[STIX]{x1D712}$ . Assertion (ii) is [Reference GrigorievGri13, Example 5.17].

For (iii), suppose that $x\in R^{\ast }(W_{g})$ is a tautological class which vanishes in $R^{\ast }(W_{g},\star )$ , and let $\unicode[STIX]{x1D70B}:E\rightarrow B$ be a fibre bundle with fibre $W_{g}$ . Then the pullback fibre bundle $\unicode[STIX]{x1D70B}^{\prime }:\unicode[STIX]{x1D70B}^{\ast }E\rightarrow E$ is a fibre bundle with fibre $W_{g}$ and canonical section, so $\unicode[STIX]{x1D70B}^{\ast }(x(\unicode[STIX]{x1D70B}))=0$ . But then

$$\begin{eqnarray}(2-2g)\cdot x(\unicode[STIX]{x1D70B})=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}})\cdot \unicode[STIX]{x1D70B}^{\ast }(x(\unicode[STIX]{x1D70B})))=0\end{eqnarray}$$

and $2-2g\neq 0$ , so $x(\unicode[STIX]{x1D70B})=0$ . This holds for any fibre bundle $\unicode[STIX]{x1D70B}$ , so $x=0\in R^{\ast }(W_{g})$ .◻

Proposition 3.3. Let $n$ be odd, let $I=(i_{1},i_{2},\ldots ,i_{n})$ be a sequence and $p_{I}=p_{1}^{i_{1}}p_{2}^{i_{2}}\cdots p_{n}^{i_{n}}$ the associated monomial in the Pontrjagin classes, and write $|I|=\sum _{j=1}^{n}i_{j}$ .

  1. (i) The class $\unicode[STIX]{x1D705}_{p_{I}}$ is nilpotent in $R^{\ast }(W_{g})$ .

  2. (ii) We have

    $$\begin{eqnarray}\unicode[STIX]{x1D712}^{|I|}\cdot \unicode[STIX]{x1D705}_{ep_{I}}=\mathop{\prod }_{j=1}^{n}\unicode[STIX]{x1D705}_{ep_{j}}^{i_{j}}\in R^{\ast }(W_{g})/\sqrt{0}.\end{eqnarray}$$
  3. (iii) If $g>1$ then the class $e$ is nilpotent in $R^{\ast }(W_{g},\star )$ .

  4. (iv) If $g\geqslant 1$ then for all $k>1$ the class $\unicode[STIX]{x1D705}_{e^{k}}$ is nilpotent in $R^{\ast }(W_{g})$ .

In Proposition 5.5 we will show that $e\in R^{\ast }(W_{1},\star )$ is also nilpotent, though the proof is very different from that of this proposition.

Proof. (i) By Theorem 2.1, the class $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}$ is nilpotent for all $i$ . Therefore, for any monomial $\tilde{{\mathcal{L}}}_{J}=\tilde{{\mathcal{L}}}_{1}^{j_{1}}\tilde{{\mathcal{L}}}_{2}^{j_{2}}\cdots \tilde{{\mathcal{L}}}_{n}^{j_{n}}$ , the class $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{J}}$ is nilpotent by Theorem 3.1. As the $\tilde{{\mathcal{L}}}_{i}$ and the $p_{i}$ generate the same subring of $H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ , it follows that any monomial $p_{I}$ may be written as a polynomial in the $\tilde{{\mathcal{L}}}_{i}$ , and hence each class $\unicode[STIX]{x1D705}_{p_{I}}$ is nilpotent.

(ii) Let $\unicode[STIX]{x1D70B}:E\rightarrow B$ be a fibre bundle with fibre $W_{g}$ , and for a monomial $p_{J}$ in the Pontrjagin classes let us write

$$\begin{eqnarray}(p_{J})^{\prime }=\unicode[STIX]{x1D712}ep_{J}-e\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D705}_{ep_{J}}).\end{eqnarray}$$

We compute that

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{(p_{J})^{\prime }p_{i}}=\unicode[STIX]{x1D70B}_{!}(\unicode[STIX]{x1D712}ep_{i}p_{J}-ep_{i}\unicode[STIX]{x1D70B}^{\ast }(\unicode[STIX]{x1D705}_{ep_{J}}))=\unicode[STIX]{x1D712}\unicode[STIX]{x1D705}_{ep_{i}p_{J}}-\unicode[STIX]{x1D705}_{ep_{i}}\unicode[STIX]{x1D705}_{ep_{J}}\end{eqnarray}$$

and that $\unicode[STIX]{x1D705}_{(p_{J})^{\prime }}=0$ because $\unicode[STIX]{x1D70B}_{!}(e)=\unicode[STIX]{x1D712}$ . By (i) the class $\unicode[STIX]{x1D705}_{p_{i}}$ is nilpotent, so by applying Theorem 3.1 it follows that $\unicode[STIX]{x1D705}_{(p_{J})^{\prime }p_{i}}$ is nilpotent. Therefore $\unicode[STIX]{x1D712}\unicode[STIX]{x1D705}_{ep_{i}p_{J}}=\unicode[STIX]{x1D705}_{ep_{i}}\unicode[STIX]{x1D705}_{ep_{J}}$ modulo nilpotents. These identities hold for any such fibre bundle, so they hold in $R^{\ast }(W_{g})$ , which by induction proves (ii).

(iii) By (i), the class $\unicode[STIX]{x1D705}_{e^{2}}=\unicode[STIX]{x1D705}_{p_{n}}$ is nilpotent, so by (2) it follows that $e\in R^{\ast }(W_{g},\star )$ is also nilpotent as long as $g>1$ .

(iv) Suppose first that $g=1$ , so that $\unicode[STIX]{x1D705}_{e}=0$ . Then, as

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{e^{2l}}=\unicode[STIX]{x1D705}_{p_{n}^{l}}\quad \text{ and }\quad \unicode[STIX]{x1D705}_{e^{2l+1}}=\unicode[STIX]{x1D705}_{ep_{n}^{l}},\end{eqnarray}$$

it follows from (i) that the former are nilpotent, and then from Theorem 3.1 that the latter are too.

For $g>1$ the class $\unicode[STIX]{x1D705}_{e}$ is not itself nilpotent. However, writing

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{e^{2l}}=\unicode[STIX]{x1D705}_{p_{n}^{l}}\quad \text{ and }\quad \unicode[STIX]{x1D705}_{e^{2l+1}}=\unicode[STIX]{x1D705}_{e^{3}p_{n}^{l-1}},\end{eqnarray}$$

we see that the same argument will go through as soon as we show that $\unicode[STIX]{x1D705}_{e^{3}}$ is nilpotent.

Applying (1) with $c=p_{n}=e^{2}$ gives

$$\begin{eqnarray}\unicode[STIX]{x1D712}^{2}p_{n}-\unicode[STIX]{x1D712}\unicode[STIX]{x1D705}_{ep_{n}}-\unicode[STIX]{x1D712}e\unicode[STIX]{x1D705}_{p_{n}}+\unicode[STIX]{x1D705}_{e^{2}}\unicode[STIX]{x1D705}_{p_{n}}=0\in R^{\ast }(M_{g},\star )/\sqrt{0}.\end{eqnarray}$$

The class $\unicode[STIX]{x1D705}_{p_{n}}$ is nilpotent by (i), so we find that $\unicode[STIX]{x1D705}_{ep_{n}}=\unicode[STIX]{x1D712}p_{n}\in R^{\ast }(W_{g},\star )/\sqrt{0}$ . By (iii) the class $p_{n}=e^{2}$ is nilpotent, and therefore so is $\unicode[STIX]{x1D705}_{ep_{n}}=\unicode[STIX]{x1D705}_{e^{3}}$ in the ring $R^{\ast }(W_{g},\star )$ . By Lemma 3.2(iii), the natural ring map $R^{\ast }(W_{g})\rightarrow R^{\ast }(W_{g},\star )$ is injective, so $\unicode[STIX]{x1D705}_{e^{3}}$ is nilpotent in $R^{\ast }(W_{g})$ too, as required.◻

Corollary 3.4. Let $n$ be odd. Then:

  1. (i) $R^{\ast }(W_{0})/\sqrt{0}$ is generated by $\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}$ ;

  2. (ii) $R^{\ast }(W_{1})/\sqrt{0}=\mathbb{Q}$ ;

  3. (iii) $R^{\ast }(W_{g})/\sqrt{0}$ is generated by $\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n-1}}$ for $g>1$ .

Proof. It follows from Proposition 3.3(i) and (ii) that for any $g$ the ring $R^{\ast }(W_{g})/\sqrt{0}$ is generated by the elements $\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}$ .

For $g=1$ we have $\unicode[STIX]{x1D705}_{e}=0$ . It follows from Theorem 3.1 and Proposition 3.3(i) that the $\unicode[STIX]{x1D705}_{ep_{i}}$ are nilpotent, and so $R^{\ast }(W_{1})/\sqrt{0}=\mathbb{Q}$ .

For $g>1$ it follows from Proposition 3.3(iv) that $\unicode[STIX]{x1D705}_{ep_{n}}=\unicode[STIX]{x1D705}_{e^{3}}$ is nilpotent. So $R^{\ast }(W_{g})/\sqrt{0}$ is generated by the elements $\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n-1}}$ .◻

Corollary 3.5. Let $n$ be odd and $c\in {\mathcal{B}}$ ; then $\unicode[STIX]{x1D712}c=\unicode[STIX]{x1D705}_{ec}\in R^{\ast }(W_{g},\star )/\sqrt{0}$ .

Proof. If $g=1$ then $\unicode[STIX]{x1D712}c=0$ , but also $\unicode[STIX]{x1D705}_{ec}$ is nilpotent by Corollary 3.4(ii). If $g\neq 1$ then $\unicode[STIX]{x1D712}\neq 0$ , and by (1) we have

$$\begin{eqnarray}\unicode[STIX]{x1D712}^{2}c-\unicode[STIX]{x1D712}\unicode[STIX]{x1D705}_{ec}-\unicode[STIX]{x1D712}e\unicode[STIX]{x1D705}_{c}+\unicode[STIX]{x1D705}_{e^{2}}\unicode[STIX]{x1D705}_{c}=0\in R^{\ast }(W_{g},\star )/\sqrt{0}.\end{eqnarray}$$

The class $\unicode[STIX]{x1D705}_{e^{2}}=\unicode[STIX]{x1D705}_{p_{n}}$ is nilpotent by Proposition 3.3(i), and the class $e$ is nilpotent by Proposition 3.3(iii); this gives the required equation.◻

Proof of Theorem 1.2(i) and (ii).

Statement (i) is Corollary 3.5. For (ii), note that if $g\neq 1$ , it follows that the map

$$\begin{eqnarray}R^{\ast }(W_{g})/\sqrt{0}\longrightarrow R^{\ast }(W_{g},\star )/\sqrt{0}\end{eqnarray}$$

is surjective, and under the same condition this map is injective by Lemma 3.2(iii).◻

In order to prove Theorem 1.1, we must show that the generators of the rings $R^{\ast }(W_{g})/\sqrt{0}$ given in Corollary 3.4 are algebraically independent. That is the subject of the next section.

4 Algebraic independence

In this section, we finish the proof of Theorem 1.1 by showing the following.

Theorem 4.1. Let $n$ be either odd or even, and set $\unicode[STIX]{x1D716}=1$ if $n$ is odd and $\unicode[STIX]{x1D716}=0$ if $n$ is even. Then:

  1. (i) the map $\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}]\rightarrow R^{\ast }(W_{0})/\sqrt{0}$ is injective;

  2. (ii) the map $\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n-\unicode[STIX]{x1D716}}}]\rightarrow R^{\ast }(W_{g})/\sqrt{0}$ is injective for $g>1$ .

When $n$ is odd, we have shown in Corollary 3.4 that these maps are both surjective, so Theorem 1.1 follows immediately. We first explain the proof of this theorem in case (i), as it is a simpler instance of the strategy we shall use to prove (ii). Consider the fibre bundle

$$\begin{eqnarray}S^{2n}\longrightarrow B\text{SO}(2n)\overset{\unicode[STIX]{x1D70B}}{\longrightarrow }B\text{SO}(2n+1),\end{eqnarray}$$

which defines a ring homomorphism $R^{\ast }(W_{0})\rightarrow H^{\ast }(B\text{SO}(2n+1);\mathbb{Q})$ . The target is the ring $\mathbb{Q}[p_{1},p_{2},\ldots ,p_{n}]$ , which contains no nilpotent elements, so the nilradical is in the kernel of this homomorphism. We may therefore consider the composition

$$\begin{eqnarray}\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}]\longrightarrow R^{\ast }(W_{0})/\sqrt{0}\overset{\unicode[STIX]{x1D719}}{\longrightarrow }H^{\ast }(B\text{SO}(2n+1);\mathbb{Q}).\end{eqnarray}$$

The fibre bundle $\unicode[STIX]{x1D70B}$ arises as the unit sphere bundle of the tautological bundle $\unicode[STIX]{x1D6FE}_{2n+1}\rightarrow B\text{SO}(2n+1)$ , so there is a bundle isomorphism $\unicode[STIX]{x1D716}^{1}\,\oplus \,T_{\unicode[STIX]{x1D70B}}\cong \unicode[STIX]{x1D70B}^{\ast }\unicode[STIX]{x1D6FE}_{2n+1}$ , and hence $p_{i}(T_{\unicode[STIX]{x1D70B}})=\unicode[STIX]{x1D70B}^{\ast }(p_{i})$ . Thus we compute

$$\begin{eqnarray}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D705}_{ep_{i}})=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}})\cdot \unicode[STIX]{x1D70B}^{\ast }(p_{i}))=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}}))\cdot p_{i}=2\cdot p_{i}\end{eqnarray}$$

(as $\unicode[STIX]{x1D712}(S^{2n})=2$ ), so the composition is injective, which proves (i).

Remark 4.2. This fibre bundle also shows that Proposition 3.3(iii) and (iv) cannot be improved in the case of $g=0$ . Indeed, the vertical tangent bundle for the fibre bundle above is easily seen to be the tautological $2n$ -dimensional bundle over $B\text{SO}(2n)$ , so that $e(T_{\unicode[STIX]{x1D70B}})=e\in H^{2n}(B\text{SO}(2n);\mathbb{Q})$ , which is not nilpotent. Furthermore, $\unicode[STIX]{x1D705}_{e^{2k+1}}=\unicode[STIX]{x1D705}_{ep_{n}^{k}}=2\cdot p_{n}^{k}\in H^{\ast }(B\text{SO}(2n+1);\mathbb{Q})$ is not nilpotent either.

Our strategy for proving (ii) will be similar, and we begin by constructing a fibre bundle with fibre $W_{g}$ and whose base is the classifying space of a Lie group.

4.1 Constructing the Lie group action

Let $G$ be the Lie group $\text{SO}(n)\times \text{SO}(n)$ . An action of $G$ on the manifold $W_{g}$ gives rise to a smooth fibre bundle $E\rightarrow BG$ with fibre $W_{g}$ . We will construct an example of such an action for which the characteristic classes $\unicode[STIX]{x1D705}_{ep_{i}}$ with $1\leqslant i\leqslant d-\unicode[STIX]{x1D716}$ are algebraically independent in $H^{\ast }(BG;\mathbb{Q})$ . In our construction, $W_{g}$ will appear as a boundary of another $G$ -invariant manifold.

The standard $n$ -dimensional representation of $\text{SO}(n)$ gives rise to two $n$ -dimensional representations of $G=\text{SO}(n)\times \text{SO}(n)$ . For $i=1,2$ , we let $V_{i}$ be the representation of $G$ where the $i$ th copy of $\text{SO}(n)$ acts in the standard way and the other copy acts trivially. Let $\mathbb{R}$ denote the trivial one-dimensional representation.

Proposition 4.3. The $G$ -representation $W=V_{1}\,\oplus \,V_{2}\,\oplus \,\mathbb{R}$ contains an embedded compact smooth manifold with boundary $H_{g}\subset W$ that has the following properties:

  1. (i) $\dim H_{g}=\dim W=2n+1$ ;

  2. (ii) $H_{g}$ is $G$ -invariant, so $\unicode[STIX]{x2202}H_{g}$ is also $G$ -invariant;

  3. (iii) $\unicode[STIX]{x2202}H_{g}$ is a smooth manifold diffeomorphic to $W_{g}$ .

Proof. We first illustrate the idea behind the construction of $H_{g}$ by presenting a construction that compromises on the smoothness conditions but is correct up to homeomorphism.

Let $B(V_{1},1)$ and $B(V_{2}\,\oplus \,\mathbb{R},1)$ denote the unit balls in the respective representations. For any $g\in \mathbb{N}$ and a sufficiently small $\unicode[STIX]{x1D700}>0$ , it is possible to embed $g$ balls of radius $\unicode[STIX]{x1D700}$ inside $B(V_{2}\,\oplus \,\mathbb{R},1)$ as disjoint and $\text{SO}(n)$ -invariant subspaces. We can thus define the following subspaces, both invariant under the $G$ -action ( $X_{g}$ is depicted in Figure 1):

$$\begin{eqnarray}\displaystyle X_{g} & = & \displaystyle B(V_{2}\,\oplus \,\mathbb{R},1)\Big\backslash\biggl(\bigsqcup _{g}B(V_{2}\,\oplus \,\mathbb{R},\unicode[STIX]{x1D700})\biggr)\subset V_{2}\,\oplus \,\mathbb{R},\nonumber\\ \displaystyle H_{g}^{\prime } & = & \displaystyle B(V_{1},1)\times X_{g}\subset V_{1}\,\oplus \,V_{2}\,\oplus \,\mathbb{R}.\nonumber\end{eqnarray}$$

The manifold $H_{g}^{\prime }\subset W$ has codimension 0 and is $G$ -invariant. Moreover, $\unicode[STIX]{x2202}H_{g}^{\prime }$ is by definition the boundary of a $(2n+1)$ -manifold obtained by attaching $g$ trivial unlinked $n$ -handles to $D^{2n+1}$ , so it is homeomorphic to  $W_{g}$ .

Figure 1. A cross-section of $X_{g}\subset \mathbb{R}^{3}$ for $(n,g)=(2,3)$ ; $\text{SO}(2)$ acts by rotating about the axis.

Figure 2. The corner $C$ and the rounded corner $S$ .

The construction so far does not prove the proposition, as $\unicode[STIX]{x2202}H_{g}^{\prime }$ is not a smooth submanifold of $W$ . We would like to smooth out the corners of the manifold $H_{g}^{\prime }$ to obtain a manifold $H_{g}$ that is again $G$ -invariant but has smooth boundary. The boundary of $H_{g}^{\prime }$ is not smooth at the set $\unicode[STIX]{x2202}B(V_{1},1)\times \unicode[STIX]{x2202}X_{g}\subset \unicode[STIX]{x2202}H_{g}$ .

The Riemannian metrics on $B(V_{1},1)$ and $X_{g}$ induced by those of $V_{1}$ and $V_{2}\,\oplus \,\mathbb{R}$ are $\text{SO}(n)$ -invariant. Choosing inward-pointing, nowhere-vanishing, $\text{SO}(n)$ -invariant vector fields on $\unicode[STIX]{x2202}B(V_{1},1)$ and $\unicode[STIX]{x2202}X_{g}$ (which may be achieved by choosing an inward-pointing, nowhere-vanishing vector field and then averaging over $\text{SO}(n)$ ) and integrating them, we can find a $G$ -invariant neighbourhood

$$\begin{eqnarray}U^{\prime }=\unicode[STIX]{x2202}B(V_{1},1)\times \unicode[STIX]{x2202}X_{g}\times C{\hookrightarrow}V_{1}\,\oplus \,V_{2}\,\oplus \,\mathbb{R}\end{eqnarray}$$

where $C=\{(x,y)\in \mathbb{R}^{2}\mid x,y\geqslant 0\}$ and has the trivial $G$ -action.

Let $S\subset C$ be a strictly convex subset of $C$ that agrees with $C$ in a complement of a compact set around the origin, but has a smooth boundary diffeomorphic to $\mathbb{R}$ (see Figure 2). We replace $U^{\prime }$ with its ( $G$ -invariant) subset $U=\unicode[STIX]{x2202}B(V_{1},1)\times \unicode[STIX]{x2202}X_{g}\times S$ to obtain $H_{g}$ . This $H_{g}$ clearly satisfies conditions (i) and (ii) of Proposition 4.3. Finally, $\unicode[STIX]{x2202}H_{g}$ is by definition the boundary of a smooth $(2n+1)$ -manifold obtained by attaching $g$ trivial unlinked $n$ -handles to $D^{2n+1}$ and smoothing corners, so it is diffeomorphic to  $W_{g}$ .◻

4.2 Proof of Theorem 4.1(ii)

Let us write

$$\begin{eqnarray}W_{g}\longrightarrow E\overset{\unicode[STIX]{x1D70B}}{\longrightarrow }B\text{SO}(n)\times B\text{SO}(n)\end{eqnarray}$$

for the fibre bundle of Proposition 4.3, which, as $H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q})$ has no nilpotent elements, gives a ring homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}:R^{\ast }(W_{g})/\sqrt{0}\longrightarrow H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q}).\end{eqnarray}$$

Theorem 4.1(ii) will follow if we show that $\unicode[STIX]{x1D719}(\unicode[STIX]{x1D705}_{ep_{1}}),\unicode[STIX]{x1D719}(\unicode[STIX]{x1D705}_{ep_{2}}),\ldots ,\unicode[STIX]{x1D719}(\unicode[STIX]{x1D705}_{ep_{n-\unicode[STIX]{x1D716}}})$ are algebraically independent.

By Proposition 4.3, the fibre bundle $\unicode[STIX]{x1D70B}$ arises as a codimension-one subbundle of the vector bundle $V_{1}\,\oplus \,V_{2}\,\oplus \,\mathbb{R}$ . We thus obtain a bundle isomorphism $\mathbb{R}\,\oplus \,T_{\unicode[STIX]{x1D70B}}\cong \unicode[STIX]{x1D70B}^{\ast }(V_{1}\,\oplus \,V_{2}\,\oplus \,\mathbb{R})$ , and hence $p_{i}(T_{\unicode[STIX]{x1D70B}})=\unicode[STIX]{x1D70B}^{\ast }(p_{i}(V_{1}\,\oplus \,V_{2}))$ . Therefore

$$\begin{eqnarray}\unicode[STIX]{x1D719}(\unicode[STIX]{x1D705}_{ep_{i}})=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}}))\cdot p_{i}(V_{1}\,\oplus \,V_{2})=\unicode[STIX]{x1D712}\cdot p_{i}(V_{1}\,\oplus \,V_{2}).\end{eqnarray}$$

As $\unicode[STIX]{x1D712}\neq 0$ , because we have assumed $g>1$ , the following lemma finishes the proof.

Lemma 4.4. The classes $p_{i}(V_{1}\,\oplus \,V_{2})$ for $1\leqslant i\leqslant n-\unicode[STIX]{x1D716}$ are algebraically independent in $H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q})$ .

Proof. Recall from [Reference Milnor and StasheffMS74, Theorem 15.21] that in terms of the Pontrjagin and Euler classes of the tautological bundle over $B\text{SO}(d)$ , we have

$$\begin{eqnarray}H^{\ast }(B\text{SO}(d);\mathbb{Q})=\left\{\begin{array}{@{}ll@{}}\mathbb{Q}[p_{1},\ldots ,p_{(d-1)/2}]\quad & \text{ for }d\text{ odd},\\ \mathbb{Q}[p_{1},\ldots ,p_{d/2-1},e]\quad & \text{ for }d\text{ even}.\end{array}\right.\end{eqnarray}$$

Moreover, if $d$ is even then $p_{d/2}=e^{2}$ . In all cases, if $i>(d-\unicode[STIX]{x1D716})/2$ then $p_{i}=0$ .

The block sum map $s:\text{SO}(n)\times \text{SO}(n)\rightarrow \text{SO}(2n)$ gives a map

$$\begin{eqnarray}Bs^{\ast }:R=H^{\ast }(B\text{SO}(2n);\mathbb{Q})\longrightarrow S=H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q})\end{eqnarray}$$

on cohomology, and the claim is that it is injective when restricted to the subalgebra $\mathbb{Q}[p_{1},p_{2},\ldots ,p_{n-\unicode[STIX]{x1D716}}]$ . As the block sum map is faithful, it follows from a theorem of Venkov [Reference VenkovVen59] that $S$ is finitely generated as a module over $R$ . By the above, $S$ is a polynomial ring on $n-1$ generators if $n$ is odd, or on $n$ generators if $n$ is even; $R$ is a polynomial algebra on $n$ generators.

We shall use the following result from commutative algebra: if $f:U\rightarrow V$ is a finite morphism between polynomial rings on the same (finite) number of generators, then it is injective. If it were not, then $\text{Ker}(f)$ would be a proper prime ideal (because $V$ , and hence $\text{Im}(f)$ , is an integral domain), so $U/\text{Ker}(f)$ would have strictly smaller Krull dimension than $U$ . But as $V$ is also finite over $U/\text{Ker}(f)$ , it too would have strictly smaller Krull dimension than $U$ , which is a contradiction.

If $n$ is even, then $Bs^{\ast }:R\rightarrow S$ is a finite morphism between polynomial rings on the same number of generators, so it is injective. If $n$ is odd, note that $e\in H^{2n}(B\text{SO}(2n);\mathbb{Q})=R$ lies in the kernel of $Bs^{\ast }$ (as $V_{1}\,\oplus \,V_{2}$ has trivial Euler class), so instead $R/(e)\rightarrow S$ is a finite morphism between polynomial rings on the same number of generators, and hence is injective. In either case, the subalgebra $\mathbb{Q}[p_{1},p_{2},\ldots ,p_{n-\unicode[STIX]{x1D716}}]$ of $H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ injects under the block sum map.◻

5 Addenda

5.1 Classifying spaces

We have defined the tautological ring $R^{\ast }(M)$ as a quotient of the abstract polynomial ring $\mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]$ , but it may also be described as a subring of the cohomology of the classifying space $B\text{Diff}^{+}(M)$ of the group of orientation-preserving diffeomorphisms of $M$ .

Precisely, if $\unicode[STIX]{x1D70B}:E\rightarrow B$ is equipped with the structure of a smooth oriented numerable fibre bundle with fibre $M^{d}$ , albeit without assuming that $E$ and $B$ are themselves oriented compact smooth manifolds, then we still have a vertical tangent bundle $T_{\unicode[STIX]{x1D70B}}$ and a cohomological pushforward $\unicode[STIX]{x1D70B}_{!}$ , so we may still define

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{c}(\unicode[STIX]{x1D70B})=\unicode[STIX]{x1D70B}_{!}(c(T_{\unicode[STIX]{x1D70B}}))\in H^{\ast }(B;\mathbb{Q}).\end{eqnarray}$$

The projection map

$$\begin{eqnarray}E\text{Diff}^{+}(M)\times _{\text{Diff}^{+}(M)}M\longrightarrow B\text{Diff}^{+}(M)\end{eqnarray}$$

is the universal example of such a fibre bundle, so there are universal classes $\unicode[STIX]{x1D705}_{c}\in H^{\ast }(B\text{Diff}^{+}(M);\mathbb{Q})$ .

Lemma 5.1. The ring homomorphism $\mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]\rightarrow H^{\ast }(B\text{Diff}^{+}(M);\mathbb{Q})$ has kernel $I_{M}$ , and hence its image is isomorphic to $R^{\ast }(M)$ .

Proof. The only point which needs discussion is that we defined $I_{M}$ in terms of the collection of all smooth fibre bundles over compact smooth oriented manifolds (in other words, proper submersions $\unicode[STIX]{x1D70B}:E^{k+d}\rightarrow B^{k}$ ), and $B\text{Diff}^{+}(M)$ is not one.

Let us write $I_{M}^{\prime }$ for the kernel of this ring homomorphism. As $B\text{Diff}^{+}(M)$ carries the universal smooth oriented fibre bundle with fibre $M$ , any class in $I_{M}^{\prime }$ must be trivial when evaluated on any proper submersion $\unicode[STIX]{x1D70B}:E^{k+d}\rightarrow B^{k}$ , so $I_{M}^{\prime }\subset I_{M}$ . The reverse inclusion holds because a rational cohomology class is zero if and only if it evaluates to zero on every rational homology class, and every rational homology class is represented by a map from a smooth (stably framed) manifold.◻

The analogous argument identifies $R^{\ast }(M,\star )$ with a subring of the cohomology of $B\text{Diff}^{+}(M,\star )$ , the classifying space of the group of orientation-preserving diffeomorphisms of $M$ which fix a point $\star \in M$ , and identifies $R^{\ast }(M,D^{d})$ with a subring of the cohomology of $B\text{Diff}^{+}(M,D^{d})$ , the classifying space of the group of orientation-preserving diffeomorphisms of $M$ which fix a disc $D^{d}\subset M$ .

5.2 Other classes of fibre bundles

We have defined the tautological ring $R^{\ast }(M)=\mathbb{Q}[\unicode[STIX]{x1D705}_{c}\mid c\in {\mathcal{B}}]/I_{M}$ in terms of the ideal $I_{M}$ of polynomials in the $\unicode[STIX]{x1D705}_{c}$ which vanish when evaluated on every smooth oriented fibre bundle with fibre $M$ . By varying this condition, we may form the following related tautological rings:

  1. (i) $R_{\text{torelli}}^{\ast }(M)$ , defined in terms of the ideal of polynomials in the $\unicode[STIX]{x1D705}_{c}$ which vanish on every smooth oriented fibre bundle with fibre $M$ and homologically trivial action of the fundamental group of the base;

  2. (ii) $R_{0}^{\ast }(M)$ , defined in terms of the ideal of polynomials in the $\unicode[STIX]{x1D705}_{c}$ which vanish on every smooth oriented fibre bundle with fibre $M$ and isotopically trivial action of the fundamental group of the base.

These are related by surjective ring homomorphisms

$$\begin{eqnarray}R^{\ast }(M)\longrightarrow R_{\text{torelli}}^{\ast }(M)\longrightarrow R_{0}^{\ast }(M).\end{eqnarray}$$

It follows from our results that for $M=W_{g}$ and $n$ odd, these maps all become isomorphisms after dividing out each ring by its nilradical: §§ 2 and 3 give relations in $R^{\ast }(W_{g})/\sqrt{0}$ , whereas § 4 shows algebraic independence of classes in $R_{0}^{\ast }(W_{g})/\sqrt{0}$ . In fact, this is true quite generally.

Proposition 5.2. Let $M$ be a simply connected manifold of dimension $d>5$ . Then

$$\begin{eqnarray}R^{\ast }(M)/\sqrt{0}\longrightarrow R_{\text{torelli}}^{\ast }(M)/\sqrt{0}\longrightarrow R_{0}^{\ast }(M)/\sqrt{0}\end{eqnarray}$$

are isomorphisms. The same holds for $(M,\star )$ and $(M,D^{d})$ .

Proof. Both maps are epimorphisms by definition, so it is enough to show that the composition is an isomorphism. Sullivan has shown [Reference SullivanSul77, Theorem 13.3] that under the stated conditions the mapping class group $\unicode[STIX]{x1D6E4}_{M}=\unicode[STIX]{x1D70B}_{0}(\text{Diff}^{+}(M))$ is commensurable to an arithmetic group. In particular, it has finite virtual cohomological dimension and hence finite $\mathbb{Q}$ -cohomological dimension.

Letting $\text{Diff}_{0}^{+}(M)$ denote the path component of the identity, there is a fibration sequence

$$\begin{eqnarray}B\text{Diff}_{0}^{+}(M)\overset{i}{\longrightarrow }B\text{Diff}^{+}(M)\overset{p}{\longrightarrow }B\unicode[STIX]{x1D6E4}_{M}\end{eqnarray}$$

and an associated Serre spectral sequence. We then proceed much as in the proof of Theorem 3.1: any class in

$$\begin{eqnarray}\text{Ker}(i^{\ast }:H^{\ast }(B\text{Diff}^{+}(M);\mathbb{Q})\rightarrow H^{\ast }(B\text{Diff}_{0}^{+}(M);\mathbb{Q}))\end{eqnarray}$$

has positive Serre filtration, so some power of it has Serre filtration beyond the $\mathbb{Q}$ -cohomological dimension of $B\unicode[STIX]{x1D6E4}_{M}$ . Such a power is therefore zero, showing that $\text{Ker}(i^{\ast })$ consists of nilpotent elements and hence that the map

$$\begin{eqnarray}i^{\ast }:H^{\ast }(B\text{Diff}^{+}(M);\mathbb{Q})/\sqrt{0}\longrightarrow H^{\ast }(B\text{Diff}_{0}^{+}(M);\mathbb{Q})/\sqrt{0}\end{eqnarray}$$

is injective. The result now follows from the discussion of classifying spaces in the previous section.

For $(M,\star )$ and $(M,D^{d})$ the argument is identical, except that one uses the relative mapping class groups of these pairs. For $M$ simply connected, the natural map $\unicode[STIX]{x1D6E4}_{(M,\star )}\rightarrow \unicode[STIX]{x1D6E4}_{M}$ is an isomorphism, as can be seen from the fibration sequence $M\rightarrow B\text{Diff}^{+}(M,\star )\rightarrow B\text{Diff}^{+}(M)$ . The fibration $\text{SO}(d)\rightarrow B\text{Diff}^{+}(M,D^{d})\rightarrow B\text{Diff}^{+}(M,\star )$ shows that $\unicode[STIX]{x1D6E4}_{(M,D^{d})}\rightarrow \unicode[STIX]{x1D6E4}_{(M,\star )}$ is onto with kernel of order at most 2. Thus $\unicode[STIX]{x1D6E4}_{(M,D^{d})}$ still has finite $\mathbb{Q}$ -cohomological dimension.◻

5.3 Genus zero

For the manifold $W_{0}=S^{2n}$ , we can improve Theorem 1.1 to say that

$$\begin{eqnarray}R^{\ast }(W_{0})=\mathbb{Q}[\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}];\end{eqnarray}$$

that is, we do not need to divide by the nilradical, nor do we need a condition on the parity of $n$ . Let $\unicode[STIX]{x1D70B}:E\rightarrow B$ be a fibre bundle with fibre $S^{2n}$ . We use the following two stronger versions of our results.

Lemma 5.3. We have $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}(\unicode[STIX]{x1D70B})=0\in H^{\ast }(B;\mathbb{Q})$ .

Proof. Observe that in the proof of Theorem 2.1, the group $\text{Aut}(H,\unicode[STIX]{x1D706})$ is trivial when $g=0$ . Therefore, the classes $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}(\unicode[STIX]{x1D70B})$ must be pulled back through the cohomology of the contractible space $B\text{Aut}(H,\unicode[STIX]{x1D706})$ .◻

Lemma 5.4. If $p,q\in H^{\ast }(E;\mathbb{Q})$ satisfy $\unicode[STIX]{x1D70B}_{!}(p)=0$ and $\unicode[STIX]{x1D70B}_{!}(q)=0$ , then $\unicode[STIX]{x1D70B}_{!}(pq)=0$ .

Proof. It follows from the Gysin sequence for $\unicode[STIX]{x1D70B}$ that there is a $p^{\prime }\in H^{\ast }(B;\mathbb{Q})$ such that $p=\unicode[STIX]{x1D70B}^{\ast }(p^{\prime })$ ; but then $\unicode[STIX]{x1D70B}_{!}(pq)=\unicode[STIX]{x1D70B}_{!}(\unicode[STIX]{x1D70B}^{\ast }(p^{\prime })q)=p^{\prime }\cdot \unicode[STIX]{x1D70B}_{!}(q)=0$ .◻

It follows from Lemma 5.3 that $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{i}}=0\in R^{\ast }(W_{0})$ , and so, by Lemma 5.4, $\unicode[STIX]{x1D705}_{\tilde{{\mathcal{L}}}_{J}}=0\in R^{\ast }(W_{0})$ for any monomial $\tilde{{\mathcal{L}}}_{J}$ . As these give a basis for the subring of $H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ generated by Pontrjagin classes, we have that $\unicode[STIX]{x1D705}_{p_{I}}=0$ too. For a fibre bundle $\unicode[STIX]{x1D70B}:E\rightarrow B$ with fibre $S^{2n}$ , by the Gysin sequence we therefore have $p_{I}=\unicode[STIX]{x1D70B}^{\ast }((p_{I})^{\prime })$ for some $(p_{I})^{\prime }\in H^{\ast }(B;\mathbb{Q})$ , and so

$$\begin{eqnarray}\unicode[STIX]{x1D705}_{ep_{I}}(\unicode[STIX]{x1D70B})=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}})\unicode[STIX]{x1D70B}^{\ast }((p_{I})^{\prime }))=\unicode[STIX]{x1D70B}_{!}(e(T_{\unicode[STIX]{x1D70B}}))\cdot (p_{I})^{\prime }=2(p_{I})^{\prime }.\end{eqnarray}$$

In particular, we may express $\unicode[STIX]{x1D705}_{ep_{I}}(\unicode[STIX]{x1D70B})$ by a universal formula in the $p_{i}^{\prime }={\textstyle \frac{1}{2}}\unicode[STIX]{x1D705}_{ep_{i}}(\unicode[STIX]{x1D70B})$ ; so $\unicode[STIX]{x1D705}_{ep_{1}},\unicode[STIX]{x1D705}_{ep_{2}},\ldots ,\unicode[STIX]{x1D705}_{ep_{n}}$ generate $R^{\ast }(W_{0})$ . They are algebraically independent in this ring as we have already shown that they are algebraically independent in $R^{\ast }(W_{0})/\sqrt{0}$ .

5.4 Genus one

We have two outstanding claims to prove in the $g=1$ case. The first is Theorem 1.2(iii), and the second is that $R^{\ast }(W_{1},\star )$ is finitely generated, so that $R^{\ast }(W_{1},D^{2n})$ is too and we can deduce Corollary 1.4 from Corollary 1.3.

5.4.1 The Euler class

We first prove the analogue of Proposition 3.3(iii) for $g=1$ .

Proposition 5.5. The class $e\in R^{\ast }(W_{1},\star )$ is nilpotent.

Our proof of this proposition will use the following result, which would seem to be quite generally useful in the study of tautological rings.

Theorem 5.6. Let $B$ be a finite CW-complex, let $\unicode[STIX]{x1D70B}:E\rightarrow B$ and $\unicode[STIX]{x1D70B}^{\prime }:E^{\prime }\rightarrow B$ be smooth fibre bundles with closed $d$ -manifold fibres, and let $f:E\rightarrow E^{\prime }$ be a map over $B$ which is a fibre homotopy equivalence. Then $f^{\ast }(e(T_{\unicode[STIX]{x1D70B}^{\prime }}))=e(T_{\unicode[STIX]{x1D70B}})\in H^{d}(E;\mathbb{Z})$ .

Proof. We refer to [Reference Becker and GottliebBG76] for background on the following, and to [Reference May and SigurdssonMS06, ch. 15] for technical details of fibrewise Spanier–Whitehead duality. For a space $X$ over $B$ , let $X_{+}=X\sqcup B$ be the associated ex-space, and let $\unicode[STIX]{x1D6F4}_{B}^{\infty }X_{+}$ denote the fibrewise suspension spectrum. By [Reference Becker and GottliebBG76, Theorem 4.7] or [Reference May and SigurdssonMS06, Theorem 15.1.1], the parametrised spectrum $\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}$ is dualisable, and in fact its dual can be made rather explicit. Choose a smooth embedding $E\subset B\times \mathbb{R}^{N}$ with tubular neighbourhood $U$ and projection $p:U\rightarrow E$ , and let $U_{B}^{+}$ denote the one-point compactification of $U$ formed fibrewise over $B$ , so there is a canonical section $s_{\infty }:B\rightarrow U_{B}^{+}$ given by the points at $\infty$ . There is then a map

$$\begin{eqnarray}\displaystyle B\times S^{N} & \longrightarrow & \displaystyle E_{+}\wedge _{B}U_{B}^{+}\nonumber\\ \displaystyle (b,x) & \longmapsto & \displaystyle \left\{\begin{array}{@{}ll@{}}s(b)\wedge _{B}s_{\infty }(b)\quad & \text{ for }(b,x)\not \in U,\\ p(b,x)\wedge _{B}(b,x)\quad & \text{ for }(b,x)\in U,\end{array}\right.\nonumber\end{eqnarray}$$

giving a map

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{E}:\unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+}\longrightarrow \unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\end{eqnarray}$$

of parametrised spectra which exhibits $\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}$ and $\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}$ as Spanier–Whitehead duals. There is therefore a complementary duality map

$$\begin{eqnarray}\hat{\unicode[STIX]{x1D707}}_{E}:\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\longrightarrow \unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+},\end{eqnarray}$$

and the composition

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+} & \overset{\unicode[STIX]{x1D707}_{E}}{\longrightarrow } & \displaystyle \unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\overset{\cong }{\longrightarrow }\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\nonumber\\ \displaystyle & \overset{\text{Id}\wedge _{B}\unicode[STIX]{x1D6E5}}{\longrightarrow } & \displaystyle \unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\nonumber\\ \displaystyle & \overset{\hat{\unicode[STIX]{x1D707}}_{E}\wedge _{B}\text{Id}}{\longrightarrow } & \displaystyle \unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+}\wedge _{B}\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\overset{\simeq }{\longrightarrow }\unicode[STIX]{x1D6F4}_{B}^{\infty }E_{+}\nonumber\end{eqnarray}$$

is a lift of the Becker–Gottlieb transfer of $\unicode[STIX]{x1D70B}$ to parametrised spectra. Dualising this map gives a map of parametrised spectra

$$\begin{eqnarray}\unicode[STIX]{x1D716}_{E}:\unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}\longrightarrow \unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+}\end{eqnarray}$$

and so, upon base changing along $B\rightarrow \{\ast \}$ and then collapsing $B$ to a point, we obtain a map of spectra

$$\begin{eqnarray}c:\unicode[STIX]{x1D6F4}^{\infty -N}U^{+}\longrightarrow \unicode[STIX]{x1D6F4}^{\infty }B_{+}\longrightarrow \unicode[STIX]{x1D6F4}^{\infty }S^{0}.\end{eqnarray}$$

On the other hand, $U$ is homeomorphic to the normal bundle $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70B}}$ of $E$ in $B\times \mathbb{R}^{N}$ , so $U^{+}$ is homeomorphic to the Thom space $\text{Th}(\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70B}})$ . The pullback along $c$ of the canonical cohomology class in $H^{0}(\unicode[STIX]{x1D6F4}^{\infty }S^{0};\mathbb{Z})$ , followed by the Thom isomorphism (as $T_{\unicode[STIX]{x1D70B}}$ , and hence $\unicode[STIX]{x1D708}_{\unicode[STIX]{x1D70B}}$ , is oriented), therefore gives a class in $H^{d}(E;\mathbb{Z})$ . It follows from the Poincaré–Hopf theorem that this is $e(T_{\unicode[STIX]{x1D70B}})$ .

The analogous construction, using a tubular neighbourhood $E^{\prime }\subset V\subset B\times \mathbb{R}^{N}$ , describes $e(T_{\unicode[STIX]{x1D70B}^{\prime }})$ . The point is that the above construction uses only fibrewise Spanier–Whitehead duality, and so if $D(f_{+}):\unicode[STIX]{x1D6F4}_{B}^{\infty -N}V_{B}^{+}\rightarrow \unicode[STIX]{x1D6F4}_{B}^{\infty -N}U_{B}^{+}$ is the dual of $f_{+}$ , then we have

$$\begin{eqnarray}\unicode[STIX]{x1D716}_{E}\circ D(f_{+})\simeq \unicode[STIX]{x1D716}_{E^{\prime }}:\unicode[STIX]{x1D6F4}_{B}^{\infty -N}V_{B}^{+}\longrightarrow \unicode[STIX]{x1D6F4}_{B}^{\infty }B_{+}.\end{eqnarray}$$

Therefore, as under the Thom isomorphism $D(f_{+})^{\ast }$ induces the map $(f^{\ast })^{-1}:H^{\ast }(E;\mathbb{Z})\rightarrow H^{\ast }(E^{\prime };\mathbb{Z})$ , we have $f^{\ast }(e(T_{\unicode[STIX]{x1D70B}^{\prime }}))=e(T_{\unicode[STIX]{x1D70B}})$ .◻

The arguments in this proof can be refined to define an Euler class of the vertical tangent bundle, $e(T_{\unicode[STIX]{x1D70B}})\in H^{d}(E;\mathbb{Z})$ , when $\unicode[STIX]{x1D70B}:E\rightarrow B$ is a fibration whose fibre is a Poincaré duality space of dimension $d$ . In particular, for such a Poincaré duality space $M^{d}$ , one can define the universal such class $e\in H^{d}(B\text{hAut}^{+}(M,\star );\mathbb{Z})$ .

Proof of Proposition 5.5.

Let $\unicode[STIX]{x1D70B}:E^{k+2n}\rightarrow B^{k}$ be a homologically trivial smooth oriented fibre bundle with fibre $W_{1}$ , and let $s:B\rightarrow E$ be a section. We will show that $s^{\ast }(e(T_{\unicode[STIX]{x1D70B}}))=0\in H^{2n}(B;\mathbb{Q})$ , so that $e=0\in R_{\text{torelli}}^{\ast }(W_{1},\star )$ . Hence, by Proposition 5.2, $e$ is nilpotent in $R^{\ast }(W_{1},\star )$ .

Under the stated assumptions, the Serre spectral sequence for $\unicode[STIX]{x1D70B}$ has a product structure and collapses, giving classes $a,b\in H^{n}(E;\mathbb{Z})$ which restrict to a basis of $H^{n}(W_{1};\mathbb{Z})$ . The obstructions to finding a lift

lie in $H^{i+1}(E;\unicode[STIX]{x1D70B}_{i}(S^{n}\times S^{n}))$ for $n<i<k+2n$ , which are all torsion groups.

Let $\cdots \rightarrow B_{2}\overset{f_{1}}{\rightarrow }B_{1}\overset{f_{0}}{\rightarrow }B_{0}=B$ be a tower of finite $k$ -dimensional CW-complexes in which each $f_{i}^{\ast }:H^{\ast }(B_{i};\mathbb{Z})\rightarrow H^{\ast }(B_{i+1};\mathbb{Z})$ annihilates all torsion but is rationally injective. (Such a tower may be formed as follows: an $n$ -torsion class $x\in H^{j}(B_{i};\mathbb{Z})$ arises via Bockstein from a class $x^{\prime }\in H^{j-1}(B_{i};\mathbb{Z}/n)$ represented by a map $g_{x}:B_{i}\rightarrow K(\mathbb{Z}/n,j-1)$ , and $B_{i+1}$ may be taken to be a $k$ -skeleton of the homotopy fibre of the product of the maps $g_{x}$ over all torsion classes $x\in H^{\ast }(B_{i};\mathbb{Z})$ .) We may pull back $\unicode[STIX]{x1D70B}$ and $s$ to obtain homologically trivial bundles $\unicode[STIX]{x1D70B}_{i}:E_{i}\rightarrow B_{i}$ with sections $s_{i}$ . All of these also have Serre spectral sequences which have a product structure and collapse, and they are connected by maps $\hat{f}_{i}:E_{i+1}\rightarrow E_{i}$ covering the  $f_{i}$ .

It follows that $\hat{f}_{i}^{\ast }$ always sends a torsion class to one of higher Serre filtration, so in particular (as the Serre spectral sequence for each $\unicode[STIX]{x1D70B}_{i}$ has three rows) the composition of three such maps annihilates any torsion class. Therefore a finite composition of such maps annihilates all obstructions to finding the dashed lift above, so for some $i$ we have a map $t:E_{i}\rightarrow S^{n}\times S^{n}$ splitting the inclusion of the fibre. It follows that $\unicode[STIX]{x1D70B}_{i}$ is fibre homotopy-equivalent to the trivial $W_{1}$ -bundle, and hence by Theorem 5.6 we have

$$\begin{eqnarray}e(T_{\unicode[STIX]{x1D70B}_{i}})=0\in H^{n}(E_{i};\mathbb{Z})\end{eqnarray}$$

and so $s_{i}^{\ast }e(T_{\unicode[STIX]{x1D70B}_{i}})=0\in H^{n}(B_{i};\mathbb{Z})$ . The maps $f_{i}^{\ast }$ were rationally injective, and so $s^{\ast }e(T_{\unicode[STIX]{x1D70B}})=0\in H^{2n}(B;\mathbb{Q})$ , as required.◻

Finally, we can prove the third part of Theorem 1.2.

Proof of Theorem 1.2(iii).

We have shown that $e$ is nilpotent in $R^{\ast }(W_{1},\star )$ . Consider the fibre bundle

$$\begin{eqnarray}S^{n}\times S^{n}\longrightarrow B\text{SO}(n)\times B\text{SO}(n)\overset{\unicode[STIX]{x1D70B}}{\longrightarrow }B\text{SO}(n+1)\times B\text{SO}(n+1)\end{eqnarray}$$

and let $\unicode[STIX]{x1D70B}^{\prime }:E\rightarrow B\text{SO}(n)\times B\text{SO}(n)$ be the fibre bundle obtained by pulling $\unicode[STIX]{x1D70B}$ back along itself. Then $\unicode[STIX]{x1D70B}^{\prime }$ has a section $s^{\prime }$ , and the ring $H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q})$ has no nilpotent elements, so there is a ring homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D719}:R^{\ast }(W_{1},\star )/\sqrt{0}\longrightarrow H^{\ast }(B\text{SO}(n)\times B\text{SO}(n);\mathbb{Q}).\end{eqnarray}$$

There is a bundle isomorphism $(s^{\prime })^{\ast }(T_{\unicode[STIX]{x1D70B}^{\prime }})\cong V_{1}\,\oplus \,V_{2}$ , so $\unicode[STIX]{x1D719}(p_{i})=p_{i}(V_{1}\,\oplus \,V_{2})$ . It follows from Lemma 4.4 that $p_{1},p_{2},\ldots ,p_{n-1}\subset R^{\ast }(W_{1},\star )/\sqrt{0}$ are algebraically independent.◻

5.4.2 Finite generation

The proof in [Reference GrigorievGri13] showing that $R^{\ast }(W_{g})$ is finitely generated does not apply to the $g\leqslant 1$ case. We have computed $R^{\ast }(W_{0})$ completely, and can observe that it is finitely generated, which leaves only the case $g=1$ to consider. In fact, we do not know whether this ring is finitely generated. However, we have the following result.

Proposition 5.7. For all $g\geqslant 0$ , the ring $R^{\ast }(W_{g},\star )$ is finitely generated.

Proof. Our argument follows [Reference GrigorievGri13, Example 5.19 and Lemma 5.20], and we shall freely use the language and notation of that paper. We work in the space ${\mathcal{M}}_{g}(\{1,2,\star \})$ with three marked points $1,2$ and $\star$ . Let $p\in H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ be even-dimensional, and define

$$\begin{eqnarray}\displaystyle a & = & \displaystyle p_{(\star )}-\unicode[STIX]{x1D708}_{(1\star )}\unicode[STIX]{x1D705}_{p}\in H^{\ast }({\mathcal{M}}_{g}(\{1,2,\star \});\mathbb{Q}),\nonumber\\ \displaystyle b & = & \displaystyle \unicode[STIX]{x1D708}_{(1\star )}-\unicode[STIX]{x1D708}_{(2\star )}\in H^{\ast }({\mathcal{M}}_{g}(\{1,2,\star \});\mathbb{Q}).\nonumber\end{eqnarray}$$

Both of these classes push forward to zero in $H^{\ast }({\mathcal{M}}_{g}(\{1,2\});\mathbb{Q})$ . According to [Reference GrigorievGri13, Theorem 2.7], we therefore have

$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \Bigl(\vphantom{\mathop{\sum }^{1}}\Bigl(\unicode[STIX]{x1D70B}_{\{1,2\}}^{\{1,2,\star \}}\Bigr)_{!}((p_{(\star )}-\unicode[STIX]{x1D708}_{(1\star )}\unicode[STIX]{x1D705}_{p})(\unicode[STIX]{x1D708}_{(1\star )}-\unicode[STIX]{x1D708}_{(2\star )}))\Bigr)^{2g+1}\nonumber\\ \displaystyle & = & \displaystyle (p_{(1)}-p_{(2)}-e_{(1)}\unicode[STIX]{x1D705}_{p}+\unicode[STIX]{x1D708}_{(12)}\unicode[STIX]{x1D705}_{p})^{2g+1}\in H^{\ast }({\mathcal{M}}_{g}(\{1,2\});\mathbb{Q}).\nonumber\end{eqnarray}$$

For any $q\in H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ , the above relation can be multiplied by $q_{(2)}$ and pushed forward to $H^{\ast }({\mathcal{M}}_{g}(1);\mathbb{Q})$ to obtain a relation that expresses $\unicode[STIX]{x1D705}_{p^{2g+1}q}$ in terms of decomposable elements in the tautological subring $R^{\ast }(W_{g},\star )\subset H^{\ast }({\mathcal{M}}_{g}(1);\mathbb{Q})$ . (Recall that this ring is generated by the $\unicode[STIX]{x1D705}_{c}$ as well as $c$ for $c\in H^{\ast }(B\text{SO}(2n);\mathbb{Q})$ .) As explained in [Reference GrigorievGri13, proof of Theorem 1.1], this is sufficient to show that the ring is finitely generated.◻

It follows that the ring $R^{\ast }(W_{g},D^{2n})$ is also finitely generated.


This project was initiated while the three authors were participating in the ‘Algebraic topology’ programme at MSRI, supported by NSF grant no. 0932078000, and continued while the authors were attending the ‘Homotopy theory, manifolds, and field theories’ programme at the Hausdorff Research Institute for Mathematics.


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Figure 0

Figure 1. A cross-section of $X_{g}\subset \mathbb{R}^{3}$ for $(n,g)=(2,3)$; $\text{SO}(2)$ acts by rotating about the axis.

Figure 1

Figure 2. The corner $C$ and the rounded corner $S$.

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