Proof. We first compute $\mathrm {Ext}^i_H(L,L)$ . One has

where the map $\theta $ sends an element $f\in H^0(\mathcal {O})$ to $f\phi -\phi f$ . Because $\phi $ is $\mathcal {C}^{\infty }$ -linear, every $f\in H^0(\mathcal {O})$ commutes with it; thus, $\theta $ is the $ 0 $ map and $Ext_H^0(L,L)\cong H^0(\mathcal {O})\cong \mathbb {C}$ . Moreover, $\mathrm {Ext}_H^0(L,L)\cong \mathrm {Ext}_H^2(L,L)$ by Serre dualityFootnote ³ and $\mathrm {Ext}_H^1(L,L)\cong H^0(K_C)\oplus H^1(\mathcal {O})$ . Thus,

$$ \begin{align*} \mathrm{Ext}_H^0(L,L)\cong \mathbb{C}, \quad \mathrm{Ext}^1_H(L,L)\cong \mathbb{C}^{4}, \quad \mathrm{Ext}_H^0(L,L)\cong \mathbb{C}. \end{align*} $$

We now compute $\mathrm {Ext}^i_H(L,M)$ . One has

Although $(L,\phi )$ and $(M,\psi )$ are not isomorphic as Higgs bundles, L and M might be as vector bundles. However, one can see that this does not change the nature of the description of the normal cone. Suppose first that $L\not \cong M$ : then $LM^{-1}$ is a nontrivial degree $ 0 $ line bundle, so it has no nonzero global sections and $\mathrm {Ext}^0_H(L,M)=\mathrm {Ext}^2_H(L,M)=0$ . Also, $\mathrm {Ext}^1_H(L,M)\cong H^0(LM^{-1}K_C)\oplus H^1(LM^{-1})\cong \mathbb {C}^{2} $ .

If $L\cong M$ , then $H^0(LM^{-1})\cong H^0(\mathcal {O})\cong \mathbb {C}$ and the map $\theta $ sends f to $\phi f - f\psi $ . Because $\phi \neq \psi $ , there are no nonzero elements in $H^0(\mathcal {O})$ that commute with the Higgs fields, $\mathrm {Ext}_H^0(L,M)\cong \mathrm {Ext}^2_H(L,M) = 0 $ as before. Then the alternate sum of the dimensions of vector spaces in the sequence yields $\mathrm {Ext}^1_H(L,M)\cong \mathbb {C}^{2}$ in both cases. As a result,

$$ \begin{align*} \mathrm{Ext}_H^0(L,M)=Ext_H^2(L,M)=0, \qquad \mathrm{Ext}^1_H(L,M)\cong \mathbb{C}^{2}. \end{align*} $$

Clearly, because $(L,\phi )$ and $(M,\psi )$ are switched by the involution, there are isomorphisms $\mathrm {Ext}^i_H(M,L)\cong \mathrm {Ext}^i_H(L,M)$ and $\mathrm {Ext}^i_H(M,M)\cong \mathrm {Ext}^i_H(L,L)$ . Then

$$ \begin{align*} \begin{array}{cl} T^0 &=\mathrm{Ext}^0_H(L,L) \oplus \mathrm{Ext}^0_H(M,M)=\mathbb{C}^2;\\ T^1 &=\mathrm{Ext}^1_H(L,L) \oplus \mathrm{Ext}^1_H(M,L)\oplus \mathrm{Ext}^1_H(L,M) \oplus \mathrm{Ext}^1_H(M,M)\cong \mathbb{C}^{12};\\ T^2 &=\mathrm{Ext}^2_H(L,L) \oplus \mathrm{Ext}^2_H(M,M)\cong\mathbb{C}^2.\\ \end{array} \end{align*} $$

This completes the first part of the proof.

To prove the second statement one needs to describe the zero locus of Yoneda square and the proof of [Reference O’Grady34, Proposition 1.4.1] applies mutatis mutandis. For ease of the reader, we sketch it in terms of Higgs extensions.

Consider the map

$$ \begin{align*}\begin{array}{cclc} \overline{\Upsilon}:& \mathrm{Ext}^1_H(M,L)\oplus \mathrm{Ext}^1_H(L,M) &\longrightarrow &\mathrm{Ext}^2_H(L,L)\cong \mathbb{C} \\ \ &(b,c) &\longmapsto &b \cup c.\\ \end{array} \end{align*} $$

Observe that, because $\mathrm {Ext}^2_H(L,M)\cong \mathrm {Ext}^2_H(M,L)=0$ , $\overline {\Upsilon }$ is the map induced by $\Upsilon $ on $\mathrm {Ext}^1_H(V,V)/(\mathrm {Ext}^1(L,L)\oplus \mathrm {Ext}^1_H(M,M))$ .

As a consequence of Serre duality, $\overline {\Upsilon }$ is a perfect pairing, so $\overline {\Upsilon }^{-1}(0)$ is a smooth quadric surface in $\mathbb {C}^4$ .

Now, the isomorphism $C_v\mathcal {R}_{Dol}^{\mathrm {GL}}\cong \Upsilon ^{-1}(0)$ is a general fact of deformation theory. One can use Luna’s slice theorem for determining the fibre: let $\mathcal {U}$ be the normal slice to $\mathcal {R}_{Dol}^{\mathrm {GL}}$ in v and $\mathcal {W}:=\mathcal {U}\cap \mathrm {GL}\Sigma $ ; then

$$ \begin{align*} (C_{\Sigma}\mathcal{R}_{Dol})_v \cong (C_{\mathcal{W}}\mathcal{U})_v. \end{align*} $$

Because $T_v\mathcal {W}\cong \mathrm {Ext}^1_H(L,L)\oplus \mathrm {Ext}^1(M,M)$ (cf. [Reference O’Grady34, Claim 1.4.12]) and $C_v\mathcal {U}$ is the cone over $T_v\mathcal {W}$ with fibre $(C_{\mathcal {W}}\mathcal {U})_v$ , the fibre of the cone is $\overline {\Upsilon }^{-1}(0)$ .

Proof. Again, the proof is similar to that of [Reference O’Grady34, Proposition 1.5.1]. We restate the idea in terms of Higgs extensions. As noticed in the previous paragraph, there are natural isomorphisms

$$ \begin{align*} \mathrm{Ext}^1_H(V,V)=\Lambda^1\otimes \mathfrak{g}, \qquad \mathrm{Ext}^2_H(V,V)^0\cong sl(2) \end{align*} $$

and the Yoneda product on $\mathrm {Ext}^1$ is the composition of the form $ \omega $ on $ \Lambda ^1 $ with the bracket of $\mathfrak {g}$ . Hence, the Yoneda square $\Upsilon : \Lambda ^1\otimes \mathfrak {g} \rightarrow sl(2)$ reads as

$$ \begin{align*} \Upsilon(\sum_i \lambda_i\otimes m_i)=\sum_{i,j}\omega(\lambda_i,\lambda_j) [m_i,m_j]. \end{align*} $$

Now if $\mathfrak {g}=gl(2)$ , then

$$ \begin{align*}\Lambda^1\otimes gl(2)\cong (\Lambda^1\otimes sl(2))\oplus (\Lambda^1\otimes \mathbb{C} \mathrm{Id}_V).\end{align*} $$

Let $\overline {\Upsilon }:=\Upsilon _{\mid \Lambda ^1\otimes \mathbb {C} \mathrm {Id}_V}$ . Thanks to the self-duality of $sl(2)$ as an algebra and to the identifications

$$ \begin{align*}sl(2)\otimes \Lambda^1\cong Hom(sl(2),\Lambda^1), \qquad sl(2)\cong \bigwedge^2 sl(2),\end{align*} $$

there is a well-defined map

$$ \begin{align*} \begin{array}{cclc} \overline{\Upsilon}: &Hom(sl(2),\Lambda^1)&\rightarrow &\bigwedge^2 sl(2)\\ &f &\mapsto &2f^*\omega\\ \end{array} \end{align*} $$

and $\overline {\Upsilon }^{-1}(0)= Hom^{\omega }(sl(2),\Lambda ^1)$ .

As before, the isomorphism $C_v\mathcal {R}_{Dol}^G\cong \Upsilon ^{-1}(0)$ is a general fact of deformation theory. The equality at the level of fibres is a consequence of Luna’s slice theorem: let $\mathcal {U}$ be the normal slice to $\mathcal {R}_{Dol}^{\mathrm {GL}}$ in v and $\mathcal {W}:=\mathcal {U}\cap \mathrm {GL}\Omega ^{\mathrm {GL}}$ . Then

$$ \begin{align*}(C_{\Omega}\mathcal{R}_{Dol}^{\mathrm{GL}})_v \cong (C_{\mathcal{W}}\mathcal{U})_v.\end{align*} $$

Because $T_v\mathcal {W}\cong \mathrm {Ext}^1_H(L,L)$ (cf. [Reference O’Grady34, Claim 1.5.14]) and $C_v\mathcal {U}$ is the cone over $T_v\mathcal {W}$ with fibre $(C_{\mathcal {W}}\mathcal {U})_v$ , $(C_{\mathcal {W}}\mathcal {U})_v\cong \overline {\Upsilon }^{-1}(0)$ .

If $\mathfrak {g}\cong sl(2)$ , then $\Upsilon =\overline {\Upsilon }$ and $C_v\mathcal {R}_{Dol}^{\mathrm {SL}}\cong (C_\Omega \mathcal {R}_{Dol}^{\mathrm {SL}})_v. $

Proof. Consider $V=L\oplus L^{-1}$ with $L\in Jac(C)$ such that $L\not \cong L^{-1}$ . We have that

$$ \begin{align*} H^0(End_0(V)\otimes K_C)= H^0(K_C)\oplus H^0(L^2K_C)\oplus H^0(L^{-2}K_C); \end{align*} $$

thus, a Higgs field $\Phi \in H^0(End_0(V)\otimes K_C)$ will be of the form

$$ \begin{align*}\Phi=\left( \begin{array}{cc} a&b\\ c&-a\\ \end{array}\right)\end{align*} $$

with $a\in H^0(K_C)$ , $b\in H^0(L^2K_C)$ , $c\in H^0(L^{-2}K_C)$ . A pair $(V,\Phi )$ is stable if and only if both L and $L^{-1}$ are not preserved by $\Phi $ ; that is, $b,c\neq 0$ . Then one needs to understand when two different Higgs fields give rise to isomorphic Higgs bundles: because the automorphism group of V is $\mathbb {C}^*$ , two Higgs pairs $(V,\Phi _1)$ and $(V,\Phi _2)$ with $\Phi _i=(a_i,b_i,c_i)$ are isomorphic if and only if

$$ \begin{align*} \Phi_1=\left( \begin{array}{cc} t&0\\ 0&t^{-1}\\ \end{array}\right)\Phi_2\left( \begin{array}{cc} t^{-1}&0\\ 0&t\\ \end{array}\right);\end{align*} $$

that is, $a_1=a_2$ , $b_1=t^2 b_2$ , $c_1=t^{-2}c_2$ . Therefore, stable Higgs pairs $(V,\Phi )$ with fixed underline vector bundle V are parametrised by

$$ \begin{align*} H^0(K_C)\times \dfrac{(H^0(L^2K)-\{0\})\times (H^0(L^{-2}K_C)-\{0\})}{\mathbb{C}^*} \cong\mathbb{C}^2\times \mathbb{C}^*. \end{align*} $$

This is an actual quotient because all of the points are semistable with respect to the action of $\mathbb {C}^*$ .

Letting V vary, one obtains a $(\mathbb {C}^2\times \mathbb {C}^*)$ -bundle $\mathcal {S}_1$ over $\mathcal {K}^0$ . Considering $\mathcal {S}_1$ as a sphere bundle over $\mathcal {K}^0$ , the cohomology of the total space is computed by Gysin sequence. Consider the inclusions

and the long exact sequence in cohomology

which splits in

Because $\mathcal {K}^0$ is not compact, $H^0_c(\mathcal {K}^0)=0$ , so $H^1_c(\mathcal {K}^0)\cong \mathbb {C}^{15}$ . By Poincaré duality,

$$ \begin{align*} H^0(\mathcal{K}^0)\cong\mathbb{C}\quad H^1(\mathcal{K}^0)=0\quad H^2(\mathcal{K}^0)=\mathbb{C}^6\quad H^3(\mathcal{K}^0)=\mathbb{C}^{15}\quad H^4(\mathcal{K}^0)=0 \end{align*} $$

with the same weights as the cohomology of $\mathcal {K}$ .

The Gysin sequence

splits in the following subsequences:

(15) $$ \begin{align} H^0(\mathcal{S}_1)\cong \mathbb{C} \quad H^3(\mathcal{S}_1)\cong\mathbb{C}^{21} \quad H^{4}(\mathcal{S}_1)\cong\mathbb{C}^{15}; \end{align} $$

(16)

(17) $$ \begin{align} H^i(\mathcal{S}_1)=0\quad \forall i\geq 5. \end{align} $$

In (16) the map $\mathbb {C}\rightarrow \mathbb {C}^6$ is the product by the Euler class of a nontrivial bundle, which is nonzero; therefore, $H^1(\mathcal {S}_1)=0$ and $H^2(\mathcal {S}_1)=\mathbb {C}^5$ . Recall that in this case both the cup product with the Euler class and the pushforward increase weights of (1,1). As a consequence of Poincaré duality, the compact support cohomology groups of $\mathcal {S}_1$ are

$$ \begin{align*} H^i_c(\mathcal{S}_1)&=0 &\forall i=0,\ldots 5 \text{ and }i=9&\quad\\ H^6_c(\mathcal{S}_1)&=\mathbb{C}^{15} & \text{ of type }(2,2)&\quad \\ H^7_c(\mathcal{S}_1)&=\mathbb{C}^{21} &\text{ of types }19(3,3)+(2,4)+(4,2)&\\ H^8_c(\mathcal{S}_1)&=\mathbb{C}^5 & \text{ of types }3(4,4)+(3,5)+(5,3)&\\ \quad H^{10}_c(\mathcal{S}_1)&=\mathbb{C} &\text{ of type }(5,5),& \end{align*} $$

and the E-polynomial of $\mathcal {S}_1$ is

$$ \begin{align*}E(\mathcal{S}_1)(u,v)=u^5v^5+u^3v^5+u^5v^3+3u^4v^4-19u^3v^3-u^2v^4-u^4v^2+15 u^2v^2.\end{align*} $$

Proof. Consider the universal line bundle $\mathcal {L}\rightarrow \mathcal {K}^0\times C$ and let $p:\mathcal {K}^0\times C \rightarrow \mathcal {K}^0$ be the projection onto the first factor. It is well known that nontrivial extensions of $\mathcal {L}$ by $\mathcal {L}^{-1}$ are parametrised by $\mathbb {P}(R^1p_{*}\mathcal {L}^2)$ : because $R^1p_{*}\mathcal {L}^2$ is a local system on $\mathcal {K}^0$ of rank 1, there exists a unique nontrivial extension up to isomorphism. Consider the universal extension bundle $\mathcal {V}$ : it fits in the short exact sequence

(18) $$ \begin{align} 0\rightarrow \mathcal{L}\rightarrow \mathcal{V} \rightarrow \mathcal{L}^{-1}\rightarrow 0 \end{align} $$

and parametrises all of the vector bundles V on C of type (ii). In order for $(\mathcal {V},\Phi )$ to be a stable Higgs bundle, the Higgs field must not preserve the subbundle $\mathcal {L}$ . By an abuse of notation, we denote by $K_C$ the pullback of the canonical bundle on C under the projection $\mathcal {K}^0\times C\rightarrow C$ . Tensoring the sequence (18) by $K_C$ and applying the covariant functor $Hom(\mathcal {V},-)$ restricted to traceless endomorphisms gives

$$ \begin{align*}0\rightarrow Hom(\mathcal{V},\mathcal{L}\otimes K_C) \rightarrow End_0(\mathcal{V})\otimes K_C \rightarrow Hom(\mathcal{V},\mathcal{L}^{-1}\otimes K_C)\rightarrow 0.\end{align*} $$

Pushing forward to $\mathcal {K}^0$ gives a long exact sequence

(19)

A Higgs pair $(\mathcal {V},\Phi )$ is stable if and only if the Higgs field $\Phi $ lies in the complement of the kernel of the restriction map $\rho : p_* End_0(\mathcal {V})\otimes K_C\rightarrow p_*\mathcal {L}^{-2}K_C$ . In order to prove the proposition, we show that $\rho =0$ .

Applying the contravariant functor $p_*Hom(-,\mathcal {L} K_C)$ to (18) gives a long exact sequence

(20)

Consider the fibre of (20) on a point $L\in \mathcal {K}^0$ . One has

Because $H^1(L^2K_C)=0$ , $H^1(V^*LK_C)\cong H^1(K_C)\cong \mathbb {C}$ ; thus, $R^1p_*Hom(\mathcal {V},\mathcal {L} K_C)$ is a local system of rank 1 on $\mathcal {K}^0\times C$ . Now consider (19) in the fibre over $L\in \mathcal {K}^0$ :

As seen before, $H^1(V^*LK_C)\cong H^1(K_C)\cong \mathbb {C}$ and $H^0(L^{-2}K_C)\cong \mathbb {C}$ , so the map ext in (19) is either 0 or an isomorphism. However, because V is a nontrivial extension, such a map has to be nonzero; thus, it is an isomorphism. As a result, $\rho $ is zero.

Proof. Up to tensor by $L\in \mathcal {K}_0$ one may restrict to the case, $L=\mathcal {O}$ , so that V is just the trivial bundle $\mathcal {O}\oplus \mathcal {O}$ . In this case $H^0(\mathcal {E}nd_0(V)\otimes K_C)\cong H^0(K_C)\otimes sl(2)\cong \mathbb {C}^2\otimes sl(2)$ and the Higgs field is of the form

$$ \begin{align*}\Phi=\left( \begin{array}{cc} a&b\\ c&-a\\ \end{array}\right) \text{ with } a,b,c\in H^0(K_C). \end{align*} $$

The bundle is not stable if and only if $\Phi $ is conjugate to an upper triangular matrix of elements of $H^0(K_C)$ . Because the action of $\mathrm {SL}(2,\mathbb {C}) $ on $H^0(K_C)\otimes sl(2)$ is trivial on the first factor, one can consider $\Phi \in H^0(K_C)\otimes sl(2) $ as a pair of matrices $(A,B)\in sl(2)\oplus sl(2)$ on which $\mathrm {SL}(2,\mathbb {C})$ acts by simultaneous conjugation. Then $\Phi $ is conjugate to an upper triangular matrix of elements of $H^0(K_C)$ if and only if A and B are simultaneously triangulable; that is, if they have a common eigenspace. By a result of Shemesh [Reference Shemesh38], this happens if and only if $\ker [A,B]\neq 0$ ; that is, $\det ([A,B])=0$ . Writing

(21) $$ \begin{align} A=\left( \begin{array}{cc} x_1&x_2\\[-2pt] x_3&-x_1\\[-2pt] \end{array}\right)\qquad B=\left( \begin{array}{cc} y_1&y_2\\[-2pt] y_3&-y_1\\[-2pt] \end{array}\right)\end{align} $$

then

$$ \begin{align*}[A,B]=\left( \begin{array}{cc} x_2y_3-y_2x_3&2(x_1y_2-x_2y_1)\\ 2(x_3y_1-x_1y_3)&-(x_2y_3-y_2x_3)\\ \end{array}\right).\end{align*} $$

One can interpret the locus of simultaneously triangulable matrices $(A,B)\in sl(2)\oplus sl(2)$ as a quartic hypersurface in $\mathbb {C}^6$ with coordinates $(x_1,x_2,x_3,y_1,y_2,y_3)$ with the equation

$$ \begin{align*}Q: (x_2y_3-y_2x_3)^2+4(x_1y_2-x_2y_1)(x_3y_1-x_1y_3)=0,\end{align*} $$

given by annihilation of $\det [A,B]$ .

Lemma 6.8. A Higgs bundle $(V,\Phi )$ with $V=\mathcal {O}\oplus \mathcal {O}$ is stable if and only if $\Phi $ lies in

$$ \begin{align*} \textbf{S}_3:=(\mathbb{C}^6-Q){\mathbin{/\mkern-5.5mu/}} \mathrm{SL}(2,\mathbb{C}), \end{align*} $$

where $\mathrm {SL}(2,\mathbb {C})$ acts as the simultaneous conjugation on the matrices A and B as in (21).

Consider the quartic hypersurface Q in $\mathbb {C}^6$ . Setting

$$ \begin{align*} \alpha &=x_2y_3-y_2x_3\\[-2pt] \beta &=x_1y_2-x_2y_1\\[-2pt] \gamma &=x_3y_1-x_1y_3, \end{align*} $$

then for every $(x_1,x_2,x_3,y_1,y_2,y_3)\in Q$ , $(\alpha ,\beta ,\gamma )$ satisfy the equation

$$ \begin{align*} \alpha^2+4\beta\gamma=0. \end{align*} $$

This defines a map $f:Q \rightarrow \mathcal {C}:=\{(\alpha ,\beta ,\gamma )\in \mathbb {C}^3\mid \alpha ^2+4\beta \gamma =0\}$ ,

$$ \begin{align*}f(x_1,x_2,x_3,y_1,y_2,y_3)= (x_2y_3-y_2x_3,x_1y_2-x_2y_1,x_3y_1-x_1y_3). \end{align*} $$

Our strategy to compute the cohomology of $(\mathbb {C}^6-Q)$ is the following:

1. 1) We decompose Q as a disjoint union of the close set $Q_0=f^{-1}(0)$ and its open complement $Q-Q_0= f^{-1}(\mathcal {C}-\{0\})$ .

2. 2) We compute the cohomology with compact support of both $Q_0$ and $Q-Q_0$ and use the additivity property to compute the cohomology with compact support of Q.

3. 3) Again, because $\mathbb {C}^6=Q\sqcup (\mathbb {C}^6-Q)$ , by the additivity property of the cohomology with compact support one can compute the cohomology of $\mathbb {C}^6-Q$ .

Observe that $\alpha ,\beta ,\gamma $ are the minors of order 2 of the matrix

(22) $$ \begin{align} \left(\begin{array}{cc} x_1 &y_1\\[-2pt] x_2 &y_2\\[-2pt] x_3 &y_3\\ \end{array} \right). \end{align} $$

Also, fixing $(x_1,x_2,x_3, y_1,y_2,y_3)$ and the corresponding point $(\alpha ,\beta ,\gamma ) \in \mathcal {C}$ , one notices that both $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ are orthogonal to $(\alpha ,\frac {\gamma }{2},\frac {\beta }{2})$ ; that is, they satisfy the equations

$$ \begin{align*} 2\alpha x_1+\gamma x_2+\beta x_3=0\qquad 2\alpha y_1+\gamma y_2+\beta y_3=0. \end{align*} $$

If $(\alpha ,\beta ,\gamma )\neq (0,0,0)$ , say, $\beta \neq 0$ , then

$$ \begin{align*} x_3=\dfrac{-2\alpha x_1-\gamma x_2}{\beta}, \qquad y_3=\dfrac{-2\alpha y_1-\gamma y_2}{\beta}. \end{align*} $$

Substituting these values in (22) and annihilating the minors of order 2 give three equations all identical to

$$ \begin{align*} x_1y_2-x_2y_1=\dfrac{\beta}{2}. \end{align*} $$

Therefore, the fibre of the map f in a point of $\mathcal {C}-\{0\}$ is a quadric in $\mathbb {C}^4$ , isomorphic to $\mathrm {SL}(2,\mathbb {C})$ . Also, $\mathcal {C}-\{0\}$ is homotopy equivalent to $\mathbb {R}\mathbb {P}^3$ ; thus, it has fundamental group $\mathbb {Z}_2$ and the monodromy outside the origin is trivial because it the same as the one described in [Reference Freitag and Kiehl16, Section 3.1]. As a result, we can compute the cohomology with compact support of $Q-Q_0=f^{-1}(\mathcal {C}-0)$ via Künneth’s formula. We have

$$ \begin{align*} \quad H^4_c(Q-Q_0)=\mathbb{C}\quad H^7_c(Q-Q_0)=\mathbb{C}^2\quad H^{10}_c(Q-Q_0)=\mathbb{C} \quad H^i_c(Q-Q_0)=0 \text{ otherwise}.\end{align*} $$

Now, we need to compute the cohomology of $Q_0$ : first observe that the condition $\alpha ,\beta ,\gamma =0$ implies that the matrix (22) has rank $\leq 1$ ; that is, $(y_1,y_2,y_3)$ is a multiple of $(x_1,x_2,x_3)$ . Thus, points in $Q_0$ are parametrised by $(\mathbb {C}^3-\{0\})\times \mathbb {C} \sqcup \{0\}\times \mathbb {C}^3$ . We observe that $Q_0$ has dimension 4 and the former is an open set in it, whereas the latter is closed. Therefore, by Proposition 6.3 one has

$$ \begin{align*}\kern-6pt\begin{array}{lll} H_c^3((\mathbb{C}^3-\{0\})\times \mathbb{C})\cong\mathbb{C} &\quad \!\!H^8_c((\mathbb{C}^3-\{0\})\times \mathbb{C})\cong \mathbb{C}\quad H^i_c((\mathbb{C}^3-\{0\})\times \mathbb{C})=0 \text{ otherwise}\\ H^6_c(\{0\}\times\mathbb{C}^3)\cong \mathbb{C} &\quad \!\!H^i_c(\{0\}\times \mathbb{C}^3)=0 \text{ otherwise}. \end{array},\end{align*} $$

Hence,

$$ \begin{align*}H^3_c(Q_0)\cong H^6_c(Q_0)\cong H^8_c(Q_0)\cong\mathbb{C}, \quad H^i_c(Q_0)=0 \text{ otherwise} .\end{align*} $$

Again, by Proposition 6.3 there is a long exact sequence

Now, $H^i_c(Q)=0$ for any $i\geq 5$ because Q is affine and from the long exact sequence one can conclude that $H^7_c (Q)\cong H^8_c(Q)\cong H^{10}_c(Q)\cong \mathbb {C}$ and $H^i_c(Q)=0$ otherwise.

Finally, from Proposition 6.3 one gets the cohomology with compact support of $\mathbb {C}^6\setminus Q$ :

$$ \begin{align*} H^8_c (\mathbb{C}^6-Q)\cong H^9_c(\mathbb{C}^6- Q)\cong H^{11}_c(\mathbb{C}^6-Q)\cong H^{12}_c(\mathbb{C}^6-Q)\cong \mathbb{C},\quad H^i_c(\mathbb{C}^6-Q)=0. \end{align*} $$

Observe that $\mathrm {SL}(2,\mathbb {C}$ ) acts on $\mathbb {C}^6-Q$ with a stabiliser that is at worst $\mathbb {Z}_2$ ; therefore, one can compute its cohomology by considering it as a fibre bundle with fibre $\mathrm {SL}(2,\mathbb {C})$ on $\mathcal {S}_3$ .

Because $\mathrm {SL}(2,\mathbb {C})$ has the same homotopy type as $S^3$ , the Gysin sequence

gives

(23) $$ \begin{align} H^0(\textbf{S}_3)\cong H^1(\textbf{S}_3)\cong\mathbb{C}\qquad H^2(\textbf{S}_3)=0 \end{align} $$

(24)

(25) $$ \begin{align} H^6(\textbf{S}_3)=0,\quad H^4(\textbf{S}_3)\cong H^8(\textbf{S}_3)\cong H^{12}(\textbf{S}_3) \end{align} $$

(26) $$ \begin{align} H^3(\textbf{S}_3)\cong H^7(\textbf{S}_3)\cong H^{11}(\textbf{S}_3)\quad H^5(\textbf{S}_3)\cong H^9(\textbf{S}_3)\quad H^6(\textbf{S}_3)\cong H^{10}(\textbf{S}_3)=0. \end{align} $$

Because $\textbf {S}_3$ is nonsingular connected but not compact, $H^{12}(\textbf {S}_3)\cong H^0_c(\textbf {S}_3)=0$ ; thus, $H^4(\textbf {S}_3)\cong H^8(\textbf {S}_3)=0$ . Therefore, from (24) we deduce that $H^3(\textbf {S}_3)\cong H^5(\textbf {S}_3)=0$ , $H^7(\textbf {S}_3)\cong H^{11}(\textbf {S}_3)=0$ and $H^9(\textbf {S}_3)=0$ and the E-polynomial of $\textbf {S}_3$ is

$$ \begin{align*} E(\textbf{S}_3)(u,v)=u^3v^3-u^2v^2.\end{align*} $$

Proof. As in type $\mathrm {(iii)}$ , it is not restrictive to assume $L\cong \mathcal {O}$ . Let V be a nontrivial extension of $\mathcal {O}$ by itself: the isomorphism classes of such bundles are parametrised by

(27) $$ \begin{align} \mathbb{P}(\mathrm{Ext}^1(\mathcal{O},\mathcal{O}))\cong \mathbb{P}^1. \end{align} $$

Thus, there exists a universal extension bundle on $\mathbb {P}^1\times C$ ,

$$ \begin{align*} 0\rightarrow \mathcal{O}\rightarrow \mathcal{V}\rightarrow\mathcal{O}\rightarrow 0. \end{align*} $$

Let $p\colon \mathbb {P}^1\times C\rightarrow \mathbb {P}^1$ be the projection: as in the type (ii) case, one can tensor the short exact sequence above by $K_C$ , apply the covariant functor $Hom(\mathcal {V},-)$ and push forward to $\mathbb {P}^1$ and end up with the long exact sequence

(28)

Stable Higgs bundles are those with Higgs field in the complement of the kernel of the map

$$ \begin{align*} p_*(End_0(\mathcal{V})\otimes K_C)\rightarrow p_*Hom(\mathcal{V},K_C) \end{align*} $$

or, equivalently, the complement of the image of $p_*Hom(\mathcal {V},K_C)$ in $p_*(End_0(\mathcal {V})\otimes K_C)$ . First notice that $p_*Hom(\mathcal {V},K_C)\cong p_*K_C$ , which is a vector bundle of rank 2, and similarly that $R^1p_*Hom(\mathcal {V},K_C)\cong R^1p_*K_C$ . Because the extension is nontrivial, the map ext is nonzero and its kernel has rank 1. Starting from (27), tensoring with $K_C$ , applying the functor $Hom(-,\mathcal {O})$ restricted to traceless endomorphisms and pushing forward to $\mathbb {P}^1$ , one obtains another long exact sequence

(29)

Notice that because $R^1p_*K_C$ has rank 1 and the map $ext$ is nonzero, the last map is surjective. Hence, the cokernel of $p_*Hom(\mathcal {V},K_C)\rightarrow p_*K_C$ has rank 1 and, consequently, $p_*Hom(\mathcal {V},K_C)$ has rank 3. Returning to (28), we conclude that $p_*End_0(V)\otimes K_C$ is a vector bundle of rank 4; thus, the locus of stable pairs is fibrewise the complement of a hyperplane.

Finally, automorphisms have to be taken into account: the group of automorphisms of a nontrivial extension of $\mathcal {O}$ by itself is the additive group $(\mathbb {C},+)\subset \mathrm {SL}(2,\mathbb {C})$ , and an element $t\in (\mathbb {C},+)$ acts on the Higgs field $\Phi $ as

$$ \begin{align*} t.\Phi=\left(\begin{array}{cc} 1&t\\ 0&1 \end{array}\right)\left(\begin{array}{cc} a&b\\ c&-a \end{array}\right)\left(\begin{array}{cc} 1&-t\\ 0&1 \end{array}\right)=\left(\begin{array}{cc} 1a+tc&b-2ta-t^2c\\ c&-a-tc \end{array}\right). \end{align*} $$

Proof. Let A be the kernel of the extension map in (28), minus the zero section: thus, A is a $\mathbb {C}^*$ -bundle over $\mathbb {P}^1$ . One can think of $p_*(End_0(\mathcal {V})\otimes K_C)-p_*Hom(\mathcal {V},K_C)$ as a vector bundle of rank 3 over A. Similarly, the kernel of the extension map of (29) gives rise to a vector bundle $\mathcal {A}$ over A of rank 1 and the map

$$ \begin{align*} p_*Hom(\mathcal{V})\rightarrow p_*End_0(\mathcal{V})\otimes K_C) \end{align*} $$

lifts to a $(\mathbb {C},+)$ -equivariant map

$$ \begin{align*} [p_*(End_0(\mathcal{V})\otimes K_C)-p_*Hom(\mathcal{V},K_C)]\rightarrow \mathcal{U} \end{align*} $$

of vector bundles over A whose kernel has rank 2. The automorphism action of $(\mathbb {C},+)$ on $\mathcal {A}$ is linear and given by $a\mapsto a+tc$ ; hence, the quotient $\mathcal {A}/\mathbb {C}$ is A itself. Because the map above is equivariant, one has that

$$ \begin{align*} [p_*(End_0(\mathcal{V})\otimes K_C)-p_*Hom(\mathcal{V},K_C)]/\mathbb{C}\rightarrow \mathcal{A}/\mathbb{C}\cong A \end{align*} $$

is a vector bundle of rank 2 over A.

Lemma (6.11) allows computation of the Betti numbers of $\textbf {S}_4$ : because $\textbf {S}_4$ is homotopy equivalent to a $\mathbb {C}^*$ -bundle on $\mathbb {P}^1$ , the Gysin sequence reads as

(30) $$ \begin{align} H^0(\textbf{S}_4) \cong H^0(\mathbb{P}^1) \cong \mathbb{C}; \end{align} $$

(31)

(32) $$ \begin{align} H^3(\textbf{S}_4) \cong H^2(\mathbb{P}^1) \cong \mathbb{C}; \end{align} $$

(33) $$ \begin{align} H^i(\textbf{S}_4)=0 \text{ for all } i=4\ldots 8. \end{align} $$

Because the central map of (31) is the cup product with the Euler class of the bundle A, which is nontrivial, it is nonzero and we have $H^1(\textbf {S}_4)=H^2(\textbf {S}_4)=0$ . By Poincaré duality, the E-polynomial of $\textbf {S}_4$ is

$$ \begin{align*} E(\textbf{S}_4)=u^4v^4-u^2v^2. \end{align*} $$

By Corollary 6.12, this completes the proof of Proposition 6.10.

Proof. Recall that $\tilde {\Omega }$ consists of 16 copies of a nonsingular hypersurface $\mathcal {G}$ in $\mathbb {P}^4$ . Therefore, its cohomology is given by

$$ \begin{align*} H^0(\tilde{\Omega})&=H^2(\tilde{\Omega})=H^4(\tilde{\Omega})=H^6(\tilde{\Omega})=\mathbb{C}^{16}. \\ H^1(\tilde{\Omega})&=H^3(\tilde{\Omega})=H^5(\tilde{\Omega})=0. \end{align*} $$

The E-polynomial of $\tilde {\Omega }$ is

$$ \begin{align*} E(\tilde{\Omega})(u,v)=16u^3v^3+16u^2v^2+16uv+16. \end{align*} $$

Proof. Observe that $\tilde {\Sigma }^{\mathrm {SL}}\setminus \tilde {\Omega }^{\mathrm {SL}}$ is a $\mathbb {P}^1$ -bundle over $\Sigma ^{\mathrm {SL}}\setminus \Omega ^{\mathrm {SL}}$ . Now

$$ \begin{align*} \Sigma^{\mathrm{SL}}\setminus \Omega^{\mathrm{SL}} \cong \left( Jac(C)\times H^0(K_C)/\mathbb{Z}_2\right)\setminus \{16 \text{ points}\}. \end{align*} $$

First we notice that $\Sigma ^{\mathrm {SL}}=(Jac(C)\times H^0(K_C)/\mathbb {Z}_2$ has the same cohomology as $\mathcal {K}$ , so

$$ \begin{align*}\begin{array}{ll} H_c^4(\Sigma^{\mathrm{SL}})\cong \mathbb{C} &\text{ of type }(2,2)\\ H_c^2(\Sigma^{\mathrm{SL}})\cong \mathbb{C}^6 &\text{ of types }4(3,3)+(2,4)+(4,2)\\ H_c^8(\Sigma^{\mathrm{SL}})\cong \mathbb{C} &\text{ of type }(4,4)\\ H_c^i(\Sigma^{\mathrm{SL}})=0& \text{ otherwise.} \end{array}\end{align*} $$

By Proposition 6.3, $\Sigma ^{\mathrm {SL}}\setminus \Omega ^{\mathrm {SL}}$ has the same cohomology groups as $\Sigma ^{\mathrm {SL}}$ except for $H^1_c(\Sigma ^{\mathrm {SL}}\setminus \Omega ^{SL})\cong \mathbb {C}^{16}$ of weight 0. One has

$$ \begin{align*} E(\tilde{\Sigma}^{\mathrm{SL}}\setminus \tilde{\Omega}^{\mathrm{SL}})(u,v)&=E(\mathbb{P}^1)E(\Sigma^{\mathrm{SL}}\setminus \Omega^{\mathrm{SL}})(u,v)\\&=(uv+1)(u^4v^4+u^2v^4+u^4v^2+4u^3v^3+u^2v^2-16)\\ &=u^5v^5+5u^4v^4+u^5v^3+u^3v^5+5u^3v^3+u^2v^4+u^4v^2+u^2v^2-16uv-16. \end{align*} $$

Proof. To compute the fibre we consider $V=L\oplus M$ with $L,M\in \mathcal {J}$ .

$$ \begin{align*} H^0(End(V)\otimes K_C)= H^0(K_C)\oplus H^0(L^{-1}MK_C)\oplus H^0(M^{-1}LK_C)\oplus H^0(K_C); \end{align*} $$

thus, a Higgs field $\Phi \in H^0(End(V)\otimes K_C)$ takes the form

$$ \begin{align*}\Phi=\left( \begin{array}{cc} a&b\\ c&d\\ \end{array}\right)\end{align*} $$

with $a,d\in H^0(K_C)$ , $b\in H^0(M^{-1}LK_C) $ , $c\in H^0(L^{-1}MK_C)$ . A pair $(V,\Phi )$ is stable if and only if both L and M are not preserved by $\Phi $ ; that is, $b,c\neq 0$ . Because the automorphisms group of V is $\mathbb {C}^*\times \mathbb {C}^*$ , two Higgs pairs $(V,\Phi _1)$ and $(V,\Phi _2)$ for $\Phi _i=(a_i,b_i,c_i,d_i)$ are isomorphic if and only if

$$ \begin{align*}\Phi_1=\left( \begin{array}{cc} t&0\\ 0&s\\ \end{array}\right)\Phi_2\left( \begin{array}{cc} t^{-1}&0\\ 0&s^{-1}\\ \end{array}\right)\end{align*} $$

that is $a_1=a_2$ , $b_1=t^{-1}s b_2$ , $c_1=s^{-1}t c_2$ , $d_1=d_2$ . Therefore, the stable Higgs pairs $(V,\Phi )$ with fixed underline vector bundle V are parametrised by