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(Edited) How can I find Euler-Poincare Index with compact support of General Linear Group over $\mathbb{R}$. For example let $A$ be a locally closed subset of a manifold $X$ then: $\chi_c(A)=\chi(R\Gamma(X;\mathbb{R}_A))=\chi(R\Gamma_c(A;\mathbb{R}_A))$

Which, in a smooth case it is the same as alternating sum of Betti numbers of de Rham cohomologies with compact support. Thank you.

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  • $\begingroup$ What do you mean by the Euler characteristic with compact support? Do you mean the Euler-Poincaré characteristic of the compactly supported de Rham complex? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Aug 5, 2010 at 16:13
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    $\begingroup$ I'm guessing that's what is meant. In which case, use Poincare duality to rewrite it as the ordinary Euler characteristic up to sign. This can be computed in the usual fashion... At this point it might help to tell us what part of the story is familiar to you (I'm addressing Karl). $\endgroup$
    – Donu Arapura
    Commented Aug 5, 2010 at 16:28
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    $\begingroup$ To expand on what Donu said: please provide some motivation (why do you want to know?) and background (what do you already know? what have you already tried?). $\endgroup$
    – Andrew Stacey
    Commented Aug 5, 2010 at 17:16

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I'm going to assume that "Euler characteristic with compact support" means
"(Euler characteristic of the one point compactification) - 1".

Let me assume that n>1.
The space in question, namely $ GL ( n,R) _ +$ , has a circle action given by any $ S ^ 1 $ subgroup of $ GL(n,R) $. This action is free on $ GL(n,R) $, and fixes the point at infinity. $ S ^ 1 $-orbits contribute zero to the euler characteristic, and the point at infinity contributes 1. So $ \chi ( GL (n,R) _ +) = 1 $, and the Euler characteristic with compact support is zero.

To make te above argument precise, you need to pick a cell decomposition of $ ( GL ( n,R)/S ^ 1 ) _ + $, and use it to construct a cell decomposition of $GL ( n,R)$. Above every n-cell of the quotient space, you put a pair of cells of $GL ( n,R) _ + $, one of dimension n and one of dimension n+1 (except for the 0-cell corresponding to the point at infinity). This might fail to be a CW-complex, but you can nevertheless compute the Euler characteristic as the alternating sum of the numbers of cells in given dimensions.

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  • $\begingroup$ For a complete answer, you should mention that $GL(0,\mathbb R)$ consists of a single point, or is empty, depending on the convention, and that $\chi(GL(1,\mathbb R)) = -2$. Note that the Euler characteristic you are using is the correct one --- it's additive on disjoint unions --- but is not a homotopy invariant. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 6, 2010 at 21:24
  • $\begingroup$ Thank you all. My idea was to use Poincare Duality for $n>1$. Then using a homotopy equivalence of $GL(n)$ and $SL(n)$. Now, since Euler characteristic of a compact Lie group $Sl(n)$ for $n>1$ is zero. We will have $chi_c(Gl(n))=0.$ Which coincides with above answers. $\endgroup$
    – Karl
    Commented Aug 7, 2010 at 10:33
  • $\begingroup$ Karl: I realize now that you had thought it through and just wanted confirmation. Sorry if my comment seemed a little blunt. I also got zero using the same process. I guess you meant to write $SO(n)$ rather than $SL(n)$. $\endgroup$
    – Donu Arapura
    Commented Aug 7, 2010 at 12:53
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The group $GL(n,\mathbb{R})$ is homotopic to $O(n)$ so these two spaces have the same Euler characteristic. For $n\geq 2$, $O(n)$ is a compact smooth manifold of positive dimension with trivial tangent bundle. Hence its Euler class is trivial, and so is its Euler characteristic.

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