I’m not sure if this is what you’re looking for, but the map $\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ given by sending $0\mapsto 1\,, 1\mapsto 0$ isn't homotopic to a homomorphism. More ...

In general you should have $H^1(\pi_1(X,x),V)\cong H^1(X,L)\,,$ where on the left we have the group cohomology of $\pi_1(X,x)$ acting on $V$ according to the representation. You will also have an ...

$z-\overline{z}=0\iff y=0\,,$ so it would seem infinitely many roots is possible.

Since $H^2(\mathfrak{g},\mathbb{R})=0$ and since $G$ is simply connected, it follows from the van Est theorems that $H^2(G,S^1)=0,$ which means that all $S^1$- extensions of $G$ are trivial. But a ...

This is a (very) partial answer: Suppose the polynomial is real and of odd degree. Then this polynomial is the characteristic polynomial of some matrix acting in an odd number of dimensions. ...

There are van Est theorems that can help with the computation of group cohomology, and in degree one it's pretty simply. One such cocycle seems to be $f(x)=x-1\,,$ and I believe all others are just ...

This condition implies that $J $ preserves the natural pairing. This is quite typical; if one puts a metric on a complex manifold one usually requires the analogous condition to hold, making it a ...

If you have a morphism $f:G\to N\,,$ then you get a $K$-extension of $G$ by pulling back the $K$-extension of $N\,.$ The morphism $f$ lifts to a morphism into $M$ if and only if this extension is ...

I'll assume the coefficients are independent of $u\,.$ There is no standard variational problem associated with this PDE for nonzero $b_i\,.$ The reason is that for a linear PDE to admit a variational ...

I'm not sure if this answers your question or not, but it's true that, for a connected manifold $X\,,$ $H^1(X)\cong H^1(\Pi_1(X),\mathcal{O})\,,$ where on the right hand side we are taking groupoid ...

This seems false. As a counterexample, let $$a_n=\frac{1}{n\cosh{(\sqrt{\ln{n}})}}$$ and let $s=0.$