# $H^{p,q}(X)$ versus $H^{q}(X, \bigwedge^p TX)$

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.

• 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? Aug 4, 2020 at 14:28
• @David E Speyer Yes Aug 4, 2020 at 14:37
• 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. Aug 4, 2020 at 15:00
• @abx You mean $3$? And so? I know they don't agree. Aug 4, 2020 at 16:05
• @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). Aug 4, 2020 at 17:37

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.

• wait I thought Hodge numbers aren't topological either. More like deformation equivalent
– user158636
Aug 10, 2020 at 18:46
• Maybe I should rewrite my first paragraph to stress lack of deformation equivalence more strongly. Aug 10, 2020 at 18:47
• @crispr See if you like this version better. Thanks for the feedback! Aug 10, 2020 at 18:56
• 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. Aug 10, 2020 at 19:44