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.

  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ @David E Speyer Yes $\endgroup$
    – dayar
    Aug 4, 2020 at 14:37
  • 4
    $\begingroup$ 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
  • 3
    $\begingroup$ @abx You mean $3$? And so? I know they don't agree. $\endgroup$
    – dayar
    Aug 4, 2020 at 16:05
  • 2
    $\begingroup$ @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


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.

  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.