# Riemann-Roch as an index theorem [closed]

I am sorry to make this a new question. I would have liked to leave a comment, but I suppose I don't have enough rep for that....

So, in the accepted answer to this question I don't understand why in the second term it is $\dim H^0(X,L\otimes \Lambda^{0,1})$ and not $L^{-1}$ with the same.thing. I suspect this is some Hodge theory thing that I am not recognizing, and it is probably completely trivial, but I would appreciate if someone could enlighten me.

Thank you.

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## closed as too localized by Cam McLeman, Daniel Moskovich, Loop Space, Tim Perutz, Pete L. ClarkNov 3 '10 at 15:17

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I think you are right. It should be $ind (\partial_L)=dim H^0 (X;L) - dim H^0 (X;L^{-1} \otimes \Lambda^{0,1})$, the second summand is dual to the cokernel of $\partial_L$ by Serre duality. – Johannes Ebert Nov 2 '10 at 8:16
I both upvoted, and voted to close. So you would have enough rep to repost this good question as a comment. – Daniel Moskovich Nov 2 '10 at 11:30
Thank you for that. I made the comment where it belonged. Since I am new to MO, I am not sure what I should do now. Should I delete this, or "vote to close"? Or just wait for the moderators to take action? Thanks. – Alfonz Nov 3 '10 at 0:08
I made a comment on the accepted answer to the other question (I also agree with you). I want to give people (especially the answerer) a little time to respond, and then I or someone else can simply edit the answer. @Alfonz: perhaps it is best for you to wait until the answer is corrected (or the situation is otherwise resolved) and then delete this question. – Pete L. Clark Nov 3 '10 at 15:24
@Pete: OK, thanks. – Alfonz Nov 3 '10 at 15:57