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If the second cohomology of a Lie algerba $g$ is $H^2(g,Z)=Z$. Then what is the second cohomology of the direct product of $n$ copies of $g$? Is it $Z^n$? Can I think if this cohomology as an integral lattice? Thanks.

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what is $Z$? If the integers, how is $\mathbb Z$ a module over a Lie algebra (presumably defined over some field)? – Mariano Suárez-Alvarez Jan 18 '13 at 20:22
I'm not sure I can think of any sensible cohomology theory where $H^2$ of a product is the product of the $H^2$s. This flies in the face of the Kuenneth formula... – user30035 Jan 18 '13 at 20:52
(If $H^1(g,k)=0$ and $H^1(g,k)=k$, the Künneth formula gives something similar to what is wanted ---I have changed $Z$ by the basefield $k$, though) – Mariano Suárez-Alvarez Jan 18 '13 at 21:08
The second $H^1$ should be an $H^0$... One day we'll be able to edit comments! – Mariano Suárez-Alvarez Jan 18 '13 at 21:09
@Mariano: I agree the question isn't formulated coherently yet (and mentions no Lie groups except in the header). But in response to your further comment, there is something you can do to edit while waiting for "one day" to arrive: copy your defective comment, then delete it, then paste the original language into a new box and edit more carefully before adding. I'm also a person who types too fast and proofreads too slowly. – Jim Humphreys Jan 19 '13 at 0:00

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