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Let $X$ be a Kahler manifold. To $X$ one can associate the cohomology groups $H^{p,q}(X)$, and $H^{(0,q)}(X, \bigwedge^p TX)$ with $TX$ being the holomorphic tangent bundle of $X$.

Is there a general relationship between these two cohomologies?

For instance, it is claimed in a footnote on page 30 in Hori, Iqbal, and Vafa - D-branes and mirror symmetry, that the Euler characters of the two agree up to a sign. It would be nice to know the origin of this.

Second we know that when $X$ is Calabi–Yau the two cohomologies match. For the more general case when $X$ is Kähler it would be good to know a precise way to measure the mismatch.

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    $\begingroup$ Just to check, you understand that $H^{p,q}(X)$ is $H^q(X, \bigwedge^q T^{\ast} X)$, where $T^{\ast} X$ is the cotangent bundle, and your question is what happens if we use the tangent bundle $TX$ instead of the cotangent bundle? $\endgroup$ Aug 4, 2020 at 14:28
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    $\begingroup$ @David E Speyer Yes $\endgroup$
    – dayar
    Aug 4, 2020 at 14:37
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    $\begingroup$ By Serre duality $H^q(X,T)= H^{n-q}(X, \Omega^1\otimes \omega)^*$, where $\omega$ is the canonical bundle, and something similar holds for other exterior powers. So Euler characteristic match up to sign in the CY case, but I don't see how they would in general. $\endgroup$ Aug 4, 2020 at 15:00
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    $\begingroup$ @abx You mean $3$? And so? I know they don't agree. $\endgroup$
    – dayar
    Aug 4, 2020 at 16:05
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    $\begingroup$ @abx The Euler characteristic of Hochschild homology is 2 (there is only something in degree 0), the Euler characteristic of Hochschild cohomology is also (up to a sign) 2 (because 1-dimensional in degree 0, and 3-dimensional in degree 1). $\endgroup$
    – pbelmans
    Aug 4, 2020 at 17:37

1 Answer 1

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One reason that $H^p(X, \bigwedge^q TX)$ will not be as well behaved as $H^p(X, \bigwedge T^{\ast} X)$ is that it is not deformation invariant, and thus not topological. In other words, if we have a connected base $B$, and a flat projective family $\mathcal{X}$ over $B$ with smooth fibers, then will have the same Hodge numbers. This will not be true for $H^q(X, \bigwedge^p TX)$. Note that, in particular, this means that $H^q(X, \bigwedge^p TX)$ is not topological, since all fibers of such a family will be diffeomorphic.

Let $X_1$ be $\mathbb{P}^2$ blown up at $3$ general points, and let $X_2$ be $\mathbb{P}^2$ blownup at $3$ collinear points. I claim that $H^0(X_1, T) \cong \mathbb{C}^2$, but $H^0(X_2, T) \cong \mathbb{C}^3$. The surfaces $X_1$ and $X_2$ are diffeomorphic, and it is easy to make a smooth flat projective family over $\mathbb{A}^1$ with most fibers $X_1$ and some fibers $X_2$. This seems like bad news.

Recall that $H^0(\mathbb{P}^2, T)$ is $8$-dimensional. Writing $z_1$, $z_2$, $z_3$ for the homogenous coordinates on $\mathbb{P}^2$, it is spanned by $z_i \tfrac{\partial}{\partial z_j}$, modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$. If we write a section of $T\mathbb{P}^2$ as $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, then this section vanishes at the point $(x_1 : x_2 : x_3)$ if and only if $(x_1, x_2, x_3)$ is an eigenvector of the matrix $A^i_j$.

Let $U_j$ be the locus of $\mathbb{P}^2$ away from the points which are blown up in $X_j$. So $U_j$ is an open subset of both $\mathbb{P}^2$ and $U_j$, and we can restrict sections of the tangent bundle from the complete surfaces to $U_j$. Since $U_j$ is $\mathbb{P}^2$ remove a codimension $2$ locus, $H^0(\mathbb{P}^2, T) \cong H^0(U_j, T)$. On the other hand, I believe that a section $\sigma$ of $H^0(U_j, T)$ will extend to a section of $H^0(X_j, T)$ if and only if, considering $\sigma$ as a section on $\mathbb{P}^2$, that section vanishes at the blown up points (see computation below).

Thus, $H^0(X_j, T)$ will be sections $\sum A^i_j z_i \tfrac{\partial}{\partial z_j}$, subject to the condition that the matrix $A$ has eigenvectors at the blown up points, and modulo the relation $\sum z_k \tfrac{\partial}{\partial z_k}=0$.

If we blow up the points $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, we are requiring that the matrix $A$ be diagonal. So $H^0(X_1, T)$ is matrices of the form $\left[ \begin{smallmatrix} \ast&0&0 \\ 0&\ast&0 \\ 0&0&\ast \\ \end{smallmatrix} \right]$ modulo scalar multiples of the identity. Three general points in $\mathbb{P}^2$ are equivalent, up to $PGL_3$, to these three points, so this is the general case.

On the other hand, if we blow up $(1:0:0)$, $(0:1:0)$ and $(1:1:0)$, then we are requiring that $(\ast:\ast:0)$ be an eigenspace of $A$. This means that we are looking at matrices of the form $\left[ \begin{smallmatrix} \lambda&0&\ast \\ 0&\lambda&\ast \\ 0&0&\ast \\ \end{smallmatrix} \right]$ , modulo scalar multiples of the identity. This vector space is one dimension larger.


I had trouble finding a reference for the claim I made about vector fields extending to the blow up iff they vanish at they blown up point, so here is a proof: The statement is local, so I'll check it in the affine plane. A vector field on $\mathbb{C}^2$ corresponds to a derivation from $\mathbb{C}[x,y]$ to itself. Namely, the vector field $f(x,y) \tfrac{\partial}{\partial x} + g(x,y) \tfrac{\partial}{\partial y}$ gives the unique derivation with $D(x) = f$ and $G(y) = g$.

Such a vector field extends to the plane blown up at $(0,0)$ if and only if the derivation maps the rings $\mathbb{C}[x,y/x]$ and $\mathbb{C}[x/y,y]$ to themselves. I'll do the computation for the first case; it is enough to compute the derivation on the generators of the ring.

We have $D(x) = f \in \mathbb{C}[x,y] \subset \mathbb{C}[x,y/x]$, so there is no problem there. We then have $$D(y/x) = \frac{x D(y) - y D(x)}{x^2} = \frac{x g(x,y) - y f(x,y)}{x^2}.$$ Writing $f(x,y) = \sum f_{ij} x^i y^j$ and $g(x,y) = \sum g_{ij} x^i y^j$, we have $$\frac{x g(x,y) - y f(x,y)}{x^2} = \frac{g_{00}}{x} - \frac{f_{00} y}{x^2} + \dots$$ where the ellipses are terms that are in $\mathbb{C}[x,y] \langle 1, y/x, y^2/x^2 \rangle \subset \mathbb{C}[x,y/x]$.

So the derivation takes $\mathbb{C}[x,y/x]$ to itself if and only if $f_{00} = g_{00} = 0$, as desired.

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    $\begingroup$ wait I thought Hodge numbers aren't topological either. More like deformation equivalent $\endgroup$
    – user158636
    Aug 10, 2020 at 18:46
  • $\begingroup$ Maybe I should rewrite my first paragraph to stress lack of deformation equivalence more strongly. $\endgroup$ Aug 10, 2020 at 18:47
  • $\begingroup$ @crispr See if you like this version better. Thanks for the feedback! $\endgroup$ Aug 10, 2020 at 18:56
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    $\begingroup$ Like @crispr, I'm confused by the first paragraph. As I read it, it's phrased in a way that seems to imply that Hodge numbers are topological invariants. But they aren't ( mathoverflow.net/questions/42744/… ). Your argument, if I understand it well, should be: while $H^p(X,\wedge^q T^*_X)$ is deformation invariant, $H^p(X,\wedge^q T_X)$ is not as well behaved because it's not deformation invariant. $\endgroup$
    – Qfwfq
    Aug 10, 2020 at 19:44

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