Euler Characteristic of General Linear Group (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.
 A: 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.
A: 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.
