In "Complex Geometry" by Huybrechts, he states the following version of the Nakano identity on page $240$:

Let $X$ be a Kähler manifold and $(E,h)$ a holomorphic hermitian vector bundle on $X$. Then if $\nabla$ is the Chern connection on $E$: \begin{align*}[\Lambda,\bar{\partial}_E]=-i(\nabla^{1,0})^*\end{align*}

In the proof he appears to say that if, over some trivialising neighbourhood $U$ of $E$ where $\nabla_E = d+A$, then $\nabla_{\check{E}} = d - A$, instead of of $d - A^T$. He also doesn't state that if $E$ is trivialised over an open set $U$, then $\bar{\partial}_E = \bar{\partial}$, which I think would simplify the proof somewhat. In fact I think we have the following:

Let $X$ be a Kähler manifold and $(E,h)$ a holomorphic hermitian vector bundle. Then if $\nabla$ is any connection on $E$: \begin{align*}[\Lambda,\bar{\partial}_E]=-i(\nabla^{1,0})^*\end{align*}

I couldn't find any other references to any other form of Nakano identity (other than the Bochner-Kodaira-Nakano formula, which I am using the Nakano identity to prove) but it seemed odd that the Chern hypothesis be used unnecessarily, whence my question:

Am I missing something, or does the stronger version of the theorem really hold?

  • $\begingroup$ It's probably too late, but I'm revising the same exam as you, I think, and am confused about the same point. I also got the dual connection is $d-A^t$ instead of $d-A$ and I'm not seeing where Huybrechts uses the fact we have the Chern connection, as, once he takes an orthonormal frame, it's not in a holomorphic frame anymore. As for simplifying the proof by trivializing $\nabla$ completely, I would be worried that it wouldn't be in an orthonormal frame ($\overline{\ast}_E$ isn't hodge star). I couldn't think of a reason why your more general theorem fails. Let me know if you find anything $\endgroup$ – DCT Jun 1 '15 at 9:02
  • $\begingroup$ @Dtseng I think I found part of the problem, I was going to post an answer once exams finish, and probably e-mail Ross about it too. The fact that not all smooth frames are holomorphic is what was causing some of the problems originally. Huybrechts refers to "Remark 4.2.5" in his book for how we can take a holomorphic frame. This is where we need the connection to be the Chern connection, since properties of the matrix A are used with the content of this remark. I don't really fully understand it, and as with much of Huybrechts... $\endgroup$ – Tom Oldfield Jun 1 '15 at 20:57
  • $\begingroup$ ... there's some non-trivial work going on in the background that makes the argument go through, here we need to do some work to show that some suitable taylor series converges in remark 4.2.5. I'm sorry this isn't more useful, but if you're still interested in a couple of weeks, check back and I'll see if I can remember what I thought the answer was and post it. $\endgroup$ – Tom Oldfield Jun 1 '15 at 20:59
  • $\begingroup$ But where in the argument did we need a holomorphic frame? He showed that $[\Lambda,\overline{\partial_E}]+i(\nabla^{1,0})^*$ is linear. Then, you can pick coordinates so that $\nabla=d+A$ where $A(x)=0$ (which is true for any connection, not just a chern connection). $\endgroup$ – DCT Jun 2 '15 at 12:57
  • $\begingroup$ The existence of holomorphic local (orthonormal up to $O(|z|^2)$) frame and the that of Chern connection are both needed to show that $A(x)=0$. You may find more information in Demailly's proof of 13.1 in $L^2$ Hodge theory and vanishing theorems. $\endgroup$ – fyemath Sep 28 '18 at 18:34

The stronger version of Nakano's identity that you have mentioned cannot be true. You wrote

"Let X be a Kähler manifold and (E,h) a holomorphic hermitian vector bundle. Then if ∇ is any connection on E:

$[\Lambda, \bar{\partial}] = -i(\nabla^{1,0})^*$"

One can get a contradiction to this more general statement in the following way. Indeed, we know the Nakano identity holds for the Chern connection, so just cook up a new connection by adding to the Chern connection a non-trivial (1,0) form with values in End(E) (or with values in the bundle of skew-hermitian endomorphisms of E with respect to $h$, if you insist on the connection being compatible with $h$). Then you see that the right-hand side of the above equation gives two different answers when applied to the Chern connection and when applied to the new connection. I hope this helps.


The answer is tentative. I feel that there might be a mistake in it (see the end of the answer)

I am stuck at the same place in the textbook too, but I think it might just be a typo in the proof, instead of some more serious work going on in the background. Let's try just plowing through some computations.

As you suggested, $\nabla_{E^{\vee}}=d-A^t=d+\bar{A}$ (where we used $A^*=-A$ from the first condition that Chern connection is hermitian). Then $$(\nabla_{E}^{1,0})^*=-\bar{*} \circ (\partial+ (\bar{A})^{1,0})\circ \bar{*}=-\bar{*} \circ (\partial+ (\overline{A^{0,1}}))\circ \bar{*}=\partial^* +(\overline{A^{0,1}})^*$$

Second Chern condition of being compatible with holomorphic structure indeed tells us that $$\overline{\partial_E}=\overline{\partial}+A^{0,1}$$and looks a bit weird since we are not in the holomorphic trivialization.

Then we compute as in the book: $$[\Lambda, \overline{\partial_E}]+i (\nabla_{E}^{1,0})^*=[\Lambda, \overline{\partial}]+i\partial^*+[\Lambda, A^{0,1}]+i(\overline{A^{0,1}})^*=[\Lambda, A^{0,1}]+i(\overline{A^{0,1}})^*$$where for the first two summands we used the usual Kahler identity. The last expression is linear and we use the trick from remark 4.2.5. to show that it in fact vanishes pointwise.

PS. In fact, it indeed seems that the first condition (hermitian connection) doesn't affect the proof at all - there will be some weird adjoint in the last term and that's all.


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