Proof. For the first bound, we write

$$\begin{align*}|\eta^\vee(x)| = | {\textstyle\int \eta(\omega) e^{2 \pi i \omega x} d \omega} | \le {\textstyle\int} | \eta| \le |B| A.\end{align*}$$

For the second bound, we integrate by parts N times. For a given $x \in {\mathbb {R}}^d$ , we choose a multi-index J with $|J| =N$ and $|x|^N \sim x^J$ . Then

$$\begin{align*}|\eta^\vee(x)| = \big\lvert {\textstyle\int \eta(\omega) e^{2 \pi i \omega x} d \omega} \big\rvert = \big\lvert{\textstyle\int \partial_J \eta (2 \pi i)^{-N} x^{-J} e^{2 \pi i \omega x} d \omega} \big\rvert \lesssim |x|^{-N} {\textstyle\int |\partial_J \eta|} \le \lvert x \rvert^{-N} \lvert B \rvert A_N.\end{align*}$$

To bound $\int |\eta ^\vee (x)| dx$ , we use the first bound when $|x| \le 1$ and the second bound when $|x| \ge 1$ .

Proof. The first two bounds follow from Lemma 2.4 by a change of variables. Alternatively, one can use the same method as in Lemma 2.4.

To bound $\int |\eta ^\vee (x)| dx$ , we use the first bound when $|x| \le 1/R$ and the second bound when $|x| \ge 1/R$ .

Proof. We have $Mf = f * \eta ^\vee $ . So $\| M f \|_{L^p} \le \| f \|_{L^p} \| \eta ^\vee \|_{L^1}$ . Now apply the bound for $\| \eta ^\vee \|_{L^1}$ from Lemma 2.5.

Proof. We can first arrange that $\eta _k(\omega ) = \eta _0(2^{-k} \omega )$ . Then the function $\eta _k$ obeys the hypotheses of Lemma 2.5 with $R = 2^k$ , with bounds that are uniform in k. Then Lemma 2.5 gives the estimate $\| \eta _k^\vee \|_{L^1} \lesssim _d 1$ .

Next, we will show that $\| \partial _j \eta _k^\vee \|_{L^1} \lesssim _d 2^k$ . This will imply $\| d \eta _k^\vee \|_{L^1} \lesssim _d 2^k$ as desired.

The Fourier transform of $\partial _j \eta _k^\vee $ is $2 \pi i \omega _j \eta _k(\omega )$ . Notice that $|\omega _j| \lesssim 2^k$ on $\operatorname {\mathrm {Ann}}_k$ . We write

$$\begin{align*}2 \pi i \omega_j \eta_k = 2^k \cdot \underbrace{ 2 \pi i \frac{\omega_j}{2^k} \eta_k }_{\psi}.\end{align*}$$

The function $\psi $ obeys the hypotheses of Lemma 2.5. Therefore, $\| \psi ^\vee \|_{L^1} \lesssim _d 1.$ And so

$$\begin{align*}\| \partial_j \eta_k^\vee \|_{L^1} = 2^k \| \psi^\vee \|_{L^1} \lesssim 2^k. \\[-37pt]\end{align*}$$

Proof. $\|P_k a \|_{L^p} = \| \eta _k^\vee * a \|_{L^p} \le \| \eta _k^\vee \|_{L^1} \| a \|_{L^p}$ . Now $\| \eta _k^\vee \|_{L^1}$ is bounded uniformly in k by Lemma 2.7.

Proof. We can see this by taking the Fourier transform on both sides. The exterior derivative d becomes pointwise multiplication by a matrix on the Fourier side. The projection operator $P_k$ becomes pointwise multiplication by the scalar $\eta _k$ . These commute.

Proof. First cover $\operatorname {\mathrm {Ann}}_k$ with $\sim 1$ balls B so that the radius of each ball is $\sim 2^k$ and the distance from each ball to the origin is also $\sim 2^k$ . Let $\psi _B$ be a partition of unity: $\sum _B \psi _B = 1$ on $\operatorname {\mathrm {Ann}}_k$ and $\psi _B$ is supported in B. Decompose $a = \sum _B a_B$ where

$$\begin{align*}\hat a_B = \psi_B \hat a.\end{align*}$$

The form $\hat a_B$ is smooth and supported in $\operatorname {\mathrm {Ann}}_k \cup \operatorname {\mathrm {Ann}}_{k-1} \cup \operatorname {\mathrm {Ann}}_{k+1}$ . Just as in the proof of Lemma 2.9, it follows that $d a_B = 0$ . Using Lemma 2.6, $ \| a_B \|_{L^p} \le C \| a \|_{L^p}$ for all $1 \le p \le \infty $ .

We will construct a primitive $\operatorname {\mathrm {Prim}}(a_B)$ for each form $a_B$ such that

• • $ d \operatorname {\mathrm {Prim}}(a_B) = a_B$ .

• • $\operatorname {\mathrm {Prim}}(a_B)$ is a Schwartz form.

• • $ \lVert \operatorname {\mathrm {Prim}}(a_B) \rVert _{L^p} \le C 2^{-k} \lVert a_B \rVert _{L^p}$ for all $1 \le p \le \infty $ , with a uniform constant C.

Finally, we define $\operatorname {\mathrm {Prim}}(a) = \sum _B \operatorname {\mathrm {Prim}}(a_B)$ . Since $\operatorname {\mathrm {Prim}}(a_B)$ has the desired properties, it follows that $\operatorname {\mathrm {Prim}}(a)$ does also.

Now we have to construct $\operatorname {\mathrm {Prim}}(a_B)$ . For ease of notation, we will abbreviate $a_B$ by a. We know that $\hat a$ is supported on B. We can choose coordinates so that $\omega _1 \sim 2^k$ on B.

We write the form a as

$$\begin{align*}\sum_I a_I(x) dx_I = \sum_{I = 1 \cup J} a_I(x) dx_1 \wedge dx_J + \sum_{1 \notin I} a_I dx_I.\end{align*}$$

We define the antiderivative $\int a_I dx_1$ via the Fourier transform by the formula

(1) $$ \begin{align} \widehat {\textstyle\int a_I dx_1} (\omega) = \frac{1}{2 \pi i \omega_1} \hat a_I(\omega). \end{align} $$

Since $\omega _1> 0$ on B, and $\hat a_I(\omega )$ is supported in B, the right-hand side is a smooth compactly supported function on Fourier space. Therefore, $\int a_I dx_1$ is a Schwartz function on ${\mathbb {R}}^d$ . From (1), we can also check that

$$\begin{align*}\frac{\partial}{\partial x_1} \left( {\textstyle\int a_I dx_1} \right) = a_I.\end{align*}$$

We can also define $\int a_I dx_1$ using definite integrals:

$$\begin{align*}\int a_I dx_1 (x_1, x_2, ..., x_d) = \int_{- \infty}^{x_1} a_I(\tilde x_1, x_2, ..., x_d) d \tilde x_1.\end{align*}$$

This definite integral formula is equivalent to (1). From the definite integral formula, it takes a little work to check that $\int a_I dx_1$ is, in fact, a Schwartz function on ${\mathbb {R}}^d$ , although it is not that difficult. In our proof, we will only need (1).)

We now define

$$\begin{align*}\operatorname{\mathrm{Prim}}(a) = \sum_{I = 1 \cup J} ({\textstyle\int} a_I dx_1) dx_J.\end{align*}$$

This is a standard construction for primitives of forms which appears in the proof of the Poincaré lemma, cf. [Reference Bott and Tu6, p. 38]. We will check that $d \operatorname {\mathrm {Prim}}(a) = a$ , following the same general method as in [Reference Bott and Tu6].

We first compute $d( \int a_I dx_1)$ :

$$\begin{align*}d({\textstyle\int} a_I dx_1) = \partial_1 ({\textstyle\int} a_I dx_1) dx_1 + \sum_{j=2}^d \partial_j ({\textstyle\int} a_I dx_1) dx_j = a_I dx_1 + \sum_{j=2}^d {\textstyle\int} \partial_j a_I dx_1.\end{align*}$$

Now,

$$\begin{align*}d \operatorname{\mathrm{Prim}}(a) = \sum_{I = 1 \cup J} d ({\textstyle\int a_I dx_1}) dx_J = \sum_{I = 1 \cup J} a_I dx_1 \wedge dx_J + \sum_{I = 1 \cup J} \sum_{j=2}^d ({\textstyle\int \partial_j a_I dx_1}) dx_j \wedge dx_J.\end{align*}$$

The first term is $\sum _{I = 1 \cup J} a_I dx_I$ . So we have to check that the second term is the rest of a. In other words, we want to show that

(2) $$ \begin{align} \sum_{I = 1 \cup J} \sum_{j=2}^d ({\textstyle\int} \partial_j a_I dx_1) dx_j \wedge dx_J = \sum_{1 \notin I'} a_{I'} dx_{I'}. \end{align} $$

Since both forms are Schwartz, it suffices to check that $\partial _1$ of both sides are equal:

(3) $$ \begin{align} \sum_{I = 1 \cup J} \sum_{j=2}^d \partial_j a_I dx_j \wedge dx_J = \sum_{1 \notin I'} \partial_1 a_{I'} dx_{I'}. \end{align} $$

Since there is no 1 in J or j or $I'$ , it suffices to check that $dx_1$ wedged with both sides are equal:

(4) $$ \begin{align} \sum_{I = 1 \cup J} \sum_{j=2}^d \partial_j a_I dx_1 \wedge dx_j \wedge dx_J = \sum_{1 \notin I'} \partial_1 a_{I'} dx_1 \wedge dx_{I'}. \end{align} $$

This, in turn, follows from $da = 0$ .

To bound $\operatorname {\mathrm {Prim}}(a)$ , the main point is that $| \frac {1}{2 \pi i \omega _1}| \sim 2^{-k}$ on the ball B. Define $\eta _B = 1 $ on B, and $0 \le \eta _B \le 1$ and with $\eta _B$ supported in a slightly larger ball $\tilde B = 1.01 B$ . We can assume that $\omega _1 \sim 2^k$ on $\tilde B$ . Then

$$\begin{align*}\frac{1}{2 \pi i \omega_1} \hat a_I(\omega) = 2^{-k} \underbrace{\frac{1}{2 \pi i} \frac{2^k}{\omega_1} \eta_B}_{\tilde \eta_B} \hat a_I (\omega).\end{align*}$$

The function $\tilde \eta _B$ is supported on $\tilde B$ , and it obeys the bounds from Lemma 2.6. The lemma tells us that

$$\begin{align*}\| {\textstyle\int} a_I dx_1 \|_{L^p} = 2^{-k} \lVert\left( \tilde \eta_B \hat a_I \right)^\vee \rVert_{L^p} \le C 2^{-k} \lVert a_I \rVert_{L^p}.\end{align*}$$

Therefore, $\lVert \operatorname {\mathrm {Prim}}(a)\rVert _{L^p} \le C 2^{-k} \lVert a \rVert _{L^p}$ as desired.

Proof. By the Plancherel theorem,

$$\begin{align*}\sum_{k \in {\mathbb{Z}}} \| P_k f \|_{L^2}^2 = \sum_{k \in {\mathbb{Z}}} \int_{{\mathbb{R}}^d} \big\lvert \widehat{P_k f} \big\rvert^2 = \sum_{k \in {\mathbb{Z}}} \int_{{\mathbb{R}}^d} \lvert\eta_k(\omega)\rvert^2 \lvert\hat f(\omega)\rvert^2 d \omega.\end{align*}$$

Now for every $\omega $ , $(1/10) \le \sum _{k \in {\mathbb {Z}}} \eta _k(\omega )^2 \le 1$ . This holds because $\sum _{k \in {\mathbb {Z}}} \eta _k(\omega ) = 1$ and each $\eta _k (\omega ) \ge 0$ , and each $\omega $ lies in the support of $\eta _k$ for at most 5 values of k. Therefore,

$$\begin{align*}\sum_{k \in {\mathbb{Z}}} \| P_k f \|_{L^2}^2 = \int_{{\mathbb{R}}^d} \bigg( \sum_{k \in {\mathbb{Z}}} \eta_k(\omega)^2 \bigg) |\hat f(\omega)|^2 d \omega \sim \int_{{\mathbb{R}}^d} |\hat f (\omega)|^2 d \omega = \int_{{\mathbb{R}}^d} |f(x)|^2 dx.\end{align*}$$

For a form $a = \sum _{I} a_I(x) dx_I$ , $P_k(a) = \sum _I P_k a_I(x) dx_I$ and $\| a \|_{L^2}^2 := \sum _I \int |a_I(x)|^2 dx$ . So the case of forms follows from the case of functions.

Proof. The Fourier support of $P_{\le k} a$ is contained in the ball $B(2^{k+1}, 0)$ . For any functions f and g, the Fourier transform of $fg$ is given by

$$\begin{align*}\widehat{fg}(\omega) = \hat f * \hat g(\omega) = \int \hat f(\tilde \omega) \hat g( \omega - \tilde \omega) d \tilde \omega.\end{align*}$$

If $\hat f$ and $\hat g$ are supported in $B(2^{k+1}, 0)$ , then $\widehat {fg}$ is supported in $B( 2\cdot 2^{k+1}, 0)$ .

This argument also applies to wedge products of forms instead of products of functions, just by writing out the components of the forms. This shows that the Fourier transform of $P_{\le k} a_1 \wedge P_{\le k} a_2$ is supported in $B(2^{k+2}, 0)$ . Now, $\eta _{\le k+3}(\omega )$ is identically 1 on this ball, and so

$$\begin{align*}P_{\le k+3} \left( P_{\le k} a_1 \wedge P_{\le k} a_2 \right) = P_{\le k} a_1 \wedge P_{\le k} a_2.\\[-42pt] \end{align*}$$

Proof. The tuple $(\beta _1, \ldots , \beta _J)$ belongs to the space $\prod _{j=1}^J \Lambda ^{d_j} {\mathbb {R}}^n$ , which is isomorphic to ${\mathbb {R}}^N$ . We can think of (each component of) $\beta _j$ as a coordinate on this space, and we can think of $R_r$ as a polynomial on this space. We let $V(R_1, \ldots , R_k)$ be the set of $\vec \beta $ where all the polynomials $R_r$ vanish.

Each $(\beta _1, \ldots , \beta _J) \in V(R_1, \ldots , R_k)$ corresponds to a homomorphism $\phi : H^*(M; {\mathbb {R}}) \rightarrow \Lambda ^* {\mathbb {R}}^n$ with $\beta _j = \phi (u_j)$ . By hypothesis, each such homomorphism is noninjective. By Poincaré duality, we have that each such homomorphism sends $u_{\operatorname {\mathrm {top}}}$ to 0. Therefore, $\beta _{\operatorname {\mathrm {top}}} = 0$ on $V(R_1, \ldots , R_k)$ .

For any set $X \subset {\mathbb {R}}^N$ , we let $I(X)$ denote the ideal of polynomials $f \in {\mathbb {R}}[\beta ]$ that vanish on X. So we see that $\beta _{\operatorname {\mathrm {top}}} \in I (V (R_1, \ldots , R_k))$ . The structure of $I( V( R_1, \ldots , R_k))$ is described by the real Nullstellensatz—cf. [Reference Marshall20, §2.3]:

Theorem 2.20 (Real Nullstellensatz)

A polynomial $f \in {\mathbb {R}}[\beta ]$ lies in $I (V (R_1, \ldots , R_k))$ if and only if there is an integer $m \ge 1$ and polynomials $g_i, h_r \in {\mathbb {R}}[\beta ]$ so that

$$\begin{align*}f^{2m} + g_1^2 + \ldots + g_s^2 = \sum_{r=1}^k h_r R_r.\end{align*}$$

By the real Nullstellensatz, we see that there is some integer m such that

$$\begin{align*}\beta_{\operatorname{\mathrm{top}}}^{2m} + g_1(\beta)^2 + \ldots + g_s(\beta)^2 = \sum_r h_r(\beta) R_r(\beta).\end{align*}$$

If we also know that $|\beta _j| \le 1$ for every j and $|R_r(\beta )| \le {\epsilon }$ for every r, then we see that

$$\begin{align*}\beta_{\operatorname{\mathrm{top}}}^{2m} \le C_M {\epsilon}.\end{align*}$$

Therefore, $| \beta _{\operatorname {\mathrm {top}}} | \le C_M {\epsilon }^{\frac {1}{2m}}$ .

Proof. We start by computing

$$\begin{align*}P_{\le k} R_r(a) (x) = \int_{{\mathbb{R}}^n} \eta_k^\vee(y) R_r(a) (x-y) dy = \int_{{\mathbb{R}}^n} \eta_k^\vee(y) dg_r (x-y) dy.\end{align*}$$

Now, we can integrate by parts to get

$$\begin{align*}\int_{{\mathbb{R}}^n} \eta_k^\vee(y) dg_r (x-y) dy = \int_{{\mathbb{R}}^n} d\eta_k^\vee(y) g_r (x-y) dy.\end{align*}$$

Taking norms and using $\| g_r \|_{L^\infty } \lesssim L^{-1}$ , we see that

$$\begin{align*}| P_{\le k} R_r(a) (x) | \lesssim L^{-1} \int | d \eta_k^\vee| \lesssim 2^k L^{-1}.\\[-42pt] \end{align*}$$

Proof. Give $S^n$ the metric of $\partial [0,1]^{n+1}$ , which is $C_0$ -bilipschitz to the round metric, and divide the face $[0,1]^n \times \{0\}$ into $d^n$ identical sub-cubes, d to a side. We map all other faces to a base point, and the sub-cubes to the sphere by a rescaling of a degree 1 map

$$\begin{align*}g:([0,1]^n,\partial[0,1]^n) \to (\partial [0,1]^{n+1},\text{pt})\end{align*}$$

whose restriction to $g^{-1}([0,1]^n \times \{0\})$ is homothetic to the identity map. The resulting map has degree $d^n$ , and its Lipschitz constant in the round metric is bounded by $C_0^2(\operatorname {\mathrm {Lip}} g)d$ .

Now, consider the map $f_d \circ f_p$ . Like $f_{pd}$ , it consists of $(pd)^n$ cubical preimages of $S^n$ , with the rest of the sphere mapped to the basepoint. However, instead of one cluster of preimages filling a whole face of $\partial [0,1]^{n+1}$ , there are $p^n$ clusters of slightly smaller preimages. We homotope $f_d \circ f_p$ to $f_{pd}$ by linearly expanding these preimages to fill the whole face. The Lipschitz constant of this homotopy is bounded by $\operatorname {\mathrm {Lip}} f_d \cdot \operatorname {\mathrm {Lip}} f_p=C_0^4(\operatorname {\mathrm {Lip}} g)^2 pd$ .

Proof of Theorem 3.2

We fix a cell structure for $M=\#_k {\mathbb {C}} P^2$ consisting of one $0$ -cell, $k 2$ -cells and a $4$ -cell. Let $\iota :[0,1]^4 \to M$ be the inclusion map of the $4$ -cell and let

$$\begin{align*}\partial=\iota|_{\partial[0,1]^4}:S^3 \to M^{(2)}=\bigvee_{i=1}^k S^2\end{align*}$$

be its attaching map. The projection of $\partial $ to each $S^2$ summand has Hopf invariant one. Notice that a map $\bigvee _{i=1}^k S^2 \to \bigvee _{i=1}^k S^2$ which sends each $S^2$ to itself with degree d extends to a map $M \to M$ of degree $d^2$ .

We prove the theorem by induction on $\ell $ . For the base of the induction, we take ${r}_1:M \to M$ to be any map whose restriction to each $2$ -cell is the map $f_2:S^2 \to S^2$ from Lemma 3.3.

For the inductive step, assume that we have constructed a $C(\ell -1) \cdot 2^{\ell -1}$ -Lipschitz map ${r}_{2^{\ell -1}}:M \to M$ whose restriction to each $2$ -cell is $f_{2^{\ell -1}}$ . To build ${r}_{2^\ell }$ , we take a $2 \times 2 \times 2 \times 2$ grid of sub-cubes inside $[0,1]^4$ , each of side length $\frac {1}{2}\cdot \frac {\ell -1}{\ell }$ , and send each of them to M via a homothetic rescaling of ${r}_{2^{\ell -1}} \circ \iota $ . Then the Lipschitz constant on each sub-cube is $C\ell \cdot 2^\ell $ .

We must now extend the map to the rest of $[0,1]^4$ , filling the space in between with the same Lipschitz constant. These gaps have width on the order of $1/\ell $ .

First, we fix some notation. Let $A \subseteq [0,1]^4$ be the complement of the $16$ open subcubes

$$\begin{align*}K_{\bar a}=(a_1,a_2,a_3,a_4)+(1/8,3/8)^4,\qquad\text{for each }\bar a=(a_1,a_2,a_3,a_4),\; a_i \in \{0,1/2\},\end{align*}$$

and fix a Lipschitz map $g:A \to M^{(2)}$ which restricts to a map homothetic to $\partial $ on each $\partial K_{\bar a}$ and to $f_2 \circ \partial $ on $\partial [0,1]^4$ . Here, we write $f_d:\bigvee _{i=1}^k S^2 \to \bigvee _{i=1}^k S^2$ for the map which induces the map from Lemma 3.3 on each wedge summand.

Now, we construct ${r}_{2^\ell }$ as follows:

• • In $[0,1]^4$ but outside of $(\frac {1}{8\ell },1-\frac {1}{8\ell })^4$ , the map is a homotopy from $f_{2^\ell } \circ \partial $ to $f_{2^{\ell -1}} \circ f_2 \circ \partial $ . Such a homotopy with domain $S^3 \times [0,1]$ can be made $C_2 \cdot 2^\ell $ -Lipschitz by Lemma 3.3, so this map is $C_3\ell \cdot 2^\ell $ -Lipschitz for some fixed constant $C_3$ .

• • In $[\frac {1}{8\ell },1-\frac {1}{8\ell }]^4$ but outside of the $16$ sub-cubes of width $2\frac {\ell -1}{\ell }$ , the map is $f_{2^{\ell -1}} \circ g \circ s_\ell $ , where $s_\ell $ is a $2\ell $ -Lipschitz piecewise linear map that sends the domain to A, as shown in Figure 2.

  

  Figure 2 Inductively assembling the map ${r}_{2^\ell }$ . The light gray regions map to $M^{(2)}$ , and the dark gray regions map to the $4$ -cell. Some regions are labeled with the restriction of ${r}_{2^\ell }$ to that region.

Then we have

$$\begin{align*}\operatorname{\mathrm{Lip}} {r}_{2^\ell}=\max\{C_2\ell \cdot 2^\ell,C_1 \cdot 2^{\ell-1} \cdot \operatorname{\mathrm{Lip}} g \cdot 2\ell,\frac{2\ell}{\ell-1}\operatorname{\mathrm{Lip}} {r}_{2^{\ell-1}}\}.\end{align*}$$

By induction, the theorem is proven with $C=\max \left\{C_2,2C_1\operatorname {\mathrm {Lip}} g,\operatorname {\mathrm {Lip}} {r}_1\right\}$ .

Proof. First suppose that $d=1$ and let $n=\dim Z$ . Then Z is a wedge of n-spheres, so the base of the induction is provided by Lemma 3.3.

Now suppose that we have proved the lemma for spaces with cells in $d-1$ dimensions, in particular for $Z^{(n-1)}$ where $\dim Z=n \geq 3$ . We start by building a metric on Z as follows. First, homothetically shrink $Z^{(n-1)}$ until the attaching maps of n-cells can be given by $1$ -Lipschitz maps from $\partial [0,1]^n$ . Then give Z the nearly Euclidean metric (as defined further down in §3.3) derived from attaching cells isometric to $[0,1]^n$ .

By Proposition 3.6, proved further down, we can also assume that ${r}_p:Z \to Z$ is cellular and Lipschitz. By applying a homotopy which is constant on the $(n-1)$ -skeleton, we can also ensure that ${r}_p$ has the following property:

Such a homotopy can be performed in several steps. First, ensure that ${r}_p$ is smooth on the preimages of the ‘middle halves’ of n-cells and that the centers of the cells are regular values. Then, by composing with a homotopy that expands a small neighborhood of the center to cover the whole cell, ensure that the preimage of each open n-cell is a disjoint union of homeomorphic copies. Then, since Z is simply connected and $n \geq 3$ , it is possible to cancel out copies of opposite orientations. The details of this purely topological argument can be found, for example, in [Reference Griffiths and Morgan14, Lemma 5.3] or [Reference White24]. Finally, we can deform this map to obtain the desired geometry.

We now construct ${r}_{p^\ell }$ and $H_\ell $ by induction on $\ell $ . Suppose we have constructed a map ${r}_{p^{\ell -1}}$ that is $C({r}_p,Z)(\ell -1)^{d-1}p^{\ell -1}$ -Lipschitz and is an extension of ${r}_{p^{\ell -1}}^{(n-1)}$ to the n-cells of Z. We will homotope ${r}_{p^{\ell -1}} \circ {r}_p$ to the desired $C({r}_p,Z)\ell ^{d-1}p^\ell $ -Lipschitz map ${r}_{p^\ell }$ .

We first apply the homotopy $H_\ell ^{(n-1)}$ to $Z^{(n-1)}$ . We extend this homotopy to a n-cell e as follows. Equip e with polar coordinates $(s,\theta )$ , with $\theta \in S^{n-1}$ and $s \in [0,1)$ , and denote the attaching map of e by $\partial _e:S^{n-1} \to Z^{(n-1)}$ . We define a homotopy $\tilde H:e \times [0,1] \to Z^{(n-1)}$ by

$$\begin{align*}\tilde H(s,\theta,t)=\begin{cases} H_\ell^{(n-1)}(\partial_e(\theta),t+2(s-1)), & s \geq 1-t/2, \\ {r}_{p^{\ell-1}} \circ {r}_p(\theta,(1-t/2)^{-1}s), & s \leq 1-t/2. \end{cases}\end{align*}$$

From this formula we see that

• • When $s=1$ , $\tilde H(s,\theta ,t)$ agrees with $H_\ell ^{(n-1)}$ .

• • At $s=1-t/2$ , $\tilde H$ is continuous since

  $$\begin{align*}H_\ell^{(n-1)}(\partial_e(\theta),t+2(s-1))=H_\ell^{(n-1)}(\partial_e(\theta),0)={r}_{p^{\ell-1}} \circ {r}_p(\theta,1).\end{align*}$$

Figure 3 Stages of the homotopy $H_\ell $ , the concatenation of $\tilde H$ and J.

At this point, $\tilde H|_{e \times \{1\}}$ has different Lipschitz constants on different regions of e, which we bound by induction on $\ell $ and d:

1. (i) On the outer half of the disk, the Lipschitz constant is

   $$\begin{align*}L_1=2\operatorname{\mathrm{Lip}} H_\ell^{(n-1)} \leq 2C'({r}_p,Z^{(n-1)})\ell^{d-2}p^\ell.\end{align*}$$

2. (ii) On the inner half, but outside $\frac {1}{2}{r}_p^{-1}(e)$ (here, $\frac {1}{2}$ refers to the homothety $(s,\theta ) \mapsto (\frac {s}{2},\theta )$ ), the Lipschitz constant is

   $$\begin{align*}L_2=2\operatorname{\mathrm{Lip}}\big({r}_{p^{\ell-1}} \circ {r}_p\big) \leq \operatorname{\mathrm{Lip}}({r}_p) \cdot 2C({r}_p,Z^{(n-1)})(\ell-1)^{d-2}p^{\ell-1}.\end{align*}$$
   This bound holds because on this subdomain, the image of ${r}_p(\theta ,s/2)$ lies in $Z^{(n-1)}$ .

3. (iii) In $\frac {1}{2}{r}_p^{-1}(e)$ , the Lipschitz constant is

   $$\begin{align*}L_3=D^{-1}\operatorname{\mathrm{Lip}} {r}_{p^{\ell-1}} \leq D^{-1}C({r}_p,Z)(\ell-1)^{d-1}p^{\ell-1},\end{align*}$$
   where D is the side length of one of the subcubes comprising $\frac {1}{2}{r}_p^{-1}(e)$ .

In the second stage $J:Z \times [0,1] \to Z$ of the homotopy, which is constant on $Z^{(n-1)}$ , we expand and shrink these three regions via a product of piecewise linear homotopies of $[0,1]$ so as to equalize the Lipschitz constants. At time $1$ , e is nearly covered by a $p \times \cdots \times p$ grid of subcubes which each map to Z via $r_{p^{\ell -1}}|_e$ composed with a homothety; the outer half of $\tilde H|_{e \times \{1\}}$ is relegated to a thin shell on the outside of the cube. We can imagine expanding every part of the domain until the Lipschitz constant is $1$ on each relevant subinterval and then shrinking the whole domain proportionally. This shows that the resulting map $J|_{t=1}$ has Lipschitz constant bounded above by

$$ \begin{align*} &pDL_3+\left(\frac{1}{2}-pD\right)L_2+\frac{1}{2}L_1 \\ &\quad\leq pC({r}_p,Z)(\ell-1)^{d-1}p^{\ell-1} + \operatorname{\mathrm{Lip}}({r}_p)C({r}_p,Z^{(n-1)})(\ell-1)^{d-2}p^{\ell-1} + C'({r}_p,Z^{(n-1)})\ell^{d-2}p^\ell \\ &\quad\leq C({r}_p,Z)\ell^{d-1}p^\ell, \end{align*} $$

where the second inequality holds as long as

$$\begin{align*}C({r}_p,Z) \geq p^{-1}\operatorname{\mathrm{Lip}}({r}_p)C({r}_p,Z^{(n-1)})+C'({r}_p,Z^{(n-1)}).\end{align*}$$

Then we set ${r}_{p^\ell }=J|_{t=1}$ and $H_\ell $ to be the concatenation of $\tilde H$ and J. By computing derivatives of $\tilde H$ and J in the space and time directions, we see that

$$\begin{align*}\operatorname{\mathrm{Lip}}(H_\ell)=\max\{L_1,L_2,L_3\},\end{align*}$$

and therefore, we can set $C'({r}_p,Z) \leq \max \{2,(pD)^{-1}\}C({r}_p,Z)$ .

Proof. We use induction on skeleta. Suppose we have constructed a complex $Y^{(k)}$ with a nearly Euclidean metric and a homotopy equivalence $f:X^{(k)} \to Y^{(k)}$ . Then for every $(k+1)$ -cell of X with attaching map $g:S^k \to X^{(k)}$ , $f \circ g$ is homotopic to a Lipschitz map $\tilde g:S^k \to Y^{(k)}$ . Then we can attach a $(k+1)$ -cell along $\tilde g$ and extend f to the $(k+1)$ -cell by combining $\tilde g$ and the homotopy.

Proof of Prop. 3.6

We start by proving a lemma:

We now make f Lipschitz, also by induction on skeleta. Clearly, $f|_{X^{(0)}}$ is Lipschitz to begin with. Now suppose that $f|_{X^{(k)}}$ is Lipschitz (notice that this is true with respect to the metric induced from $X^{(k+1)}$ as well as that on $X^{(k)}$ ) and consider a $(k+1)$ -cell not in A with an inclusion map $e:D^{k+1} \to X$ . Now take a triangulation of $D^{k+1}$ at a small enough scale that $f \circ e$ takes every simplex into a Lipschitz contractible neighborhood. By induction on the skeleta of this triangulation, we deform $f \circ e$ to a Lipschitz map, while leaving it constant on $\partial D^{k+1}$ .

If f is cellular, then we can construct the $(k+1)$ st stage of the homotopy as a map to $Y^{(k+1)}$ , rather than to Y. Then the resulting map is still cellular.

Proof. We construct Z and f by induction on skeleta. We set $Z^{(0)}=Z^{(1)}=*$ . Now suppose we have built $Z^{(n)}$ and a map $f_n:Z^{(n)} \to Y$ which induces an isomorphism on $H_k({-};\mathbb {Q})$ , $k \leq n$ . By the rational relative Hurewicz theorem, the Hurewicz map induces an isomorphism

$$\begin{align*}\pi_{n+1}(Y,Z^{(n)}) \otimes \mathbb{Q} \to H_{n+1}(Y,Z^{(n)};\mathbb{Q}) \cong H_{n+1}(Y;\mathbb{Q}).\end{align*}$$

So choose elements $\alpha _1,\ldots ,\alpha _r \in \pi _{n+1}(Y,Z^{(n)})$ forming a basis for $\pi _{n+1}(Y,Z^{(n)}) \otimes \mathbb {Q}$ . We build $Z^{(n+1)}$ by attaching an $(n+1)$ -cell $e_i$ along each $\partial \alpha _i$ , $i=1,\ldots ,r$ , and extend $f_n$ to $f_{n+1}:Z^{(n+1)} \to Y$ by mapping each $e_i$ to Y via a representative of $\alpha _i$ .

Since $(f_n)_*:H_n(Z^{(n)};\mathbb {Q}) \to H_n(Y;\mathbb {Q})$ is an isomorphism, by the long exact sequence of that pair, the Hurewicz image of each $\partial \alpha _i$ is zero. Therefore, the map

$$\begin{align*}H_{n+1}(Z^{(n+1)};\mathbb{Q}) \to H_{n+1}(Z^{(n+1)},Z^{(n)};\mathbb{Q})\end{align*}$$

is an isomorphism; in other words, the cells of $Z^{(n+1)}$ form a basis for $H_{n+1}(Z^{(n+1)};\mathbb {Q})$ . Moreover, by the definition of the extension $f_{n+1}$ , the map

$$\begin{align*}(f_{n+1})_*:\pi_{n+1}(Z^{(n+1)},Z^{(n)}) \otimes \mathbb{Q} \to \pi_{n+1}(Y,Z^{(n)}) \otimes \mathbb{Q}\end{align*}$$

is an isomorphism. But these groups are naturally isomorphic to $H_{n+1}(Z^{(n+1)};\mathbb {Q})$ and $H_{n+1}(Y;\mathbb {Q})$ , respectively. This shows that $f_{n+1}$ induces a bijection on $H_{n+1}({-};\mathbb {Q})$ as well.

Once we have done this in every dimension in which $H_*(Y;\mathbb {Q}) \neq 0$ , we have constructed the desired Z. To satisfy condition (ii), notice that we can always pick the $\alpha _i$ to be integer multiples of the elements of our chosen basis.

Proof of Theorem 3.1

Let Y be a simply connected formal compact Riemannian manifold. Using Proposition 3.10, we can find a complex Z such that the cellular chain complex of Z has zero differential and a rational equivalence $g:Z \to Y$ . Moreover, by Proposition 3.9, there is a rational equivalence $f:Y \to Z$ such that $f \circ g$ induces multiplication by $a^n$ on $H^n(Y;\mathbb Q)$ , for some $a>0$ .

By Corollary 3.7, we can put a nearly Euclidean metric on Z, and by Proposition 3.5, we can assume f and g are Lipschitz.

Now let ${r}_p:Z \to Z$ be a map that induces multiplication by $p^n$ on $H^n(Z;\mathbb Q)$ . By Lemma 3.4 and Proposition 3.5, for any nearly Euclidean metric on Z, and for every $\ell $ , there are $O(p^\ell \ell ^{d-1})$ -Lipschitz maps ${r}_{p^\ell }$ homotopic to ${r}_p^\ell $ . Then the maps $f \circ {r}_{p^\ell } \circ g$ are again $O(p^\ell \ell ^{d-1})$ -Lipschitz and induce multiplication by $(ap^\ell )^n$ on $H^n(Y;\mathbb R)$ .

Proof sketch and remarks

(i) $\iff $ (ii). Since the minimal model $\mathcal M_Y^* \to \Omega ^*(Y)$ is a quasi-isomorphism, one is quasi-isomorphic to $H^*(Y;{\mathbb {R}})$ if and only if the other is. Moreover, it is a property of minimal DGAs that if $\mathcal M_Y^*$ is quasi-isomorphic to another DGA $\mathcal A$ , then there is, in fact, a quasi-isomorphism $\mathcal M_Y^* \to \mathcal A$ .

It follows from (ii) that, while many rational homotopy types may have the same cohomology ring, exactly one of these is formal, and its minimal DGA can be constructed ‘formally’ from the cohomology ring: at stage k, one adds indecomposables in degree k that kill the relative $(k+1)$ st cohomology of the map $\mu _{k-1}:\mathcal {M}_Y^*(k-1) \to H^*(Y;{\mathbb {R}})$ and extend $\mu _{k-1}$ to a map $\mu _k:\mathcal {M}_Y^*(k) \to H^*(Y;{\mathbb {R}})$ . This is the genesis of the term ‘formal’.

Using this construction, one inductively proves that (ii) $\Rightarrow $ (iii), by showing that for each $\mathcal {M}_Y^*(k)$ , $W_0$ contains cycles representing the cohomology through dimension k. For $\mathcal {M}_Y^*(2)$ , this is clearly true since $\mathcal {M}_Y^*(2) \subseteq W_0$ . Now suppose we have a map $\mu _{k-1}:\mathcal {M}_Y^*(k-1) \to H^*(Y;{\mathbb {R}})$ . By induction, the map

$$\begin{align*}(\mu_{k-1})_*:H^{k+1}(\mathcal M_Y^*(k-1)) \to H^{k+1}(Y;{\mathbb{R}})\end{align*}$$

has image in the subring generated by $H^{\leq k-1}(Y;{\mathbb {R}})$ , and therefore, we can pick preimages in $W_0$ . The rest of $H^{k+1}(\mathcal {M}_Y^*(k-1))$ is killed by differentials of elements of $V_k$ . However, the cokernel of

$$\begin{align*}(\mu_{k-1})_*:H^k(\mathcal M_Y^*(k-1)) \to H^k(Y;{\mathbb{R}}) \cong H^k(\mathcal M_Y^*(k))\end{align*}$$

is spanned by elements of $V_k$ with zero differential, which are also in $W_0$ . Together, these span $H^k(\mathcal M_Y^*)$ .

(iii) $\Rightarrow $ (iv) is also proved by induction on dimension of indecomposables. Suppose that we have defined the bigrading on $\mathcal M_Y^*(k-1)$ . By induction, the space of $(k+1)$ -cycles in $\mathcal M_Y^*(k-1)$ splits as a direct sum of subspaces $Z_i \subseteq W_i$ since differentials of terms in $W_i$ can only cancel out with those of others in $W_i$ . Moreover, all of $\bigoplus _{i \geq 1} Z_i$ must be in the image of the differential on $V_k$ . This allows us to split $V_k$ as a direct sum of elements of various $W_i$ , $i \geq 0$ , so as to ensure $dW_i \subseteq W_{i-1}$ . We assign $\ker d \subseteq V_k$ to $W_0$ .

(iv) $\Rightarrow $ (ii). If a bigrading as in (iv) exists, then $\mathcal M_Y^* \to W_0/dW_1$ is a quotient map of DGAs.

(iv) $\Rightarrow $ (v). We can define $\hat \rho _t(a)=t^{n+i}a$ for every $a \in W_i \cap V_n$ .

(v) $\Rightarrow $ (iv) (see also [Reference Sullivan23, Thm. 12.7]). This argument requires some information about the automorphism group of $\mathcal M_Y^*$ , which is a linear algebraic subgroup of the group $\bigoplus _n \operatorname {\mathrm {Aut}}(V_n)$ . Taking the ‘diagonal part’ in the Iwasawa decomposition of $\hat \rho _t$ , we get another automorphism, also inducing the map $\rho _t$ on cohomology, which has a basis of eigenvectors in each of the $V_n$ . An inductive argument then shows that these eigenvalues are of the form $t^{n+i}$ , and setting $W_i \cap V_n$ to be the eigenspace for the eigenvalue $t^{n+i}$ gives a bigrading as in (iv).