2 votes
Accepted

Show that $\mathrm{PSL}_2(C)$ is complex algebraic

The Nullstellensatz says that, if we have something that we know is a variety, then it is defined by all the equations that it satisfies—so let's just try to write down lots of obvious equations, and ...
LSpice's user avatar
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1 vote
Accepted

Connecting homomorphism in non-abelian cohomology

$\newcommand{\diag}{{\rm diag}} \newcommand{\sH}{{\mathcal H}} \newcommand{\R}{{\mathbb R}} $No, the kernel $\ker \big[H^1(\R,\mu)\to H^1(\R,G)\big]$ does not have to be a subgroup of the abelian ...
Mikhail Borovoi's user avatar
1 vote
Accepted

An explicit matrix form

It looks like you are working with respect to the orthogonal form with matrix $\begin{pmatrix} & w_0 \\ w_0 \end{pmatrix}$, where $w_0 = \operatorname{antidiag}(1, \dotsc, 1)$. That's the one ...
LSpice's user avatar
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