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For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).

Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.

Added (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem.

Theorem. Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi_1(G)\neq 0, \mathbb{Z}$. Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.

Proof. Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi_1(U(n))=\mathbb{Z}$ wheareas $\pi_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption. That is, $U(n)\not\simeq G\times M$ as $\pi_1(U(n))\not\simeq \pi_1(G)\times \pi_1(M)$. $\square$

This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.

Added (7/9/2010): I don't know why I didn't mention it before, but replacing "unitary" with "complex" in the theorem above gives the same result for e.g. complex connected semisimple Lie groups, by an identical proof. In this case the manifold $M$ constructed in the proof cannot be guaranteed to be compact however.

For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).

Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.

Added (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem.

Theorem. Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi_1(G)\neq 0, \mathbb{Z}$. Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.

Proof. Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi_1(U(n))=\mathbb{Z}$ wheareas $\pi_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption. That is, $U(n)\not\simeq G\times M$ as $\pi_1(U(n))\not\simeq \pi_1(G)\times \pi_1(M)$. $\square$

This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.

3 Edited my answer to reflect edits in the question.

For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).

Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.

As Dona Arapura suggests, your question is not well-defined at the moment for non-Abelian $G$.

2 deleted 2 characters in body
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