Let $G$ be a reductive group defined over an algebraically closed field $k$ and $B$ be a fixed Borel subgroup of $G$. Suppose $X=Spec(R)$ is an affine scheme with $B$ rationally acting on it; hence $B$ acts on $R$. Then one can construct an induced $G$-algebra as follow: $$ S=(k[G]\otimes R)^B. $$ Now, is it true that the spectrum of $S$ is the associated scheme $G\times^BX$?
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If $X=pt$, then $G\times^B X=G/B$ is the flag variety, which is not affine.
However we have $S=\Gamma(G\times^B X, \mathcal O)$ and hence $spec(S)$ is the affinization of $G\times^B X$.
answered Nov 5, 2013 at 17:41
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$\begingroup$ Ah, you got there first... $\endgroup$ Commented Nov 5, 2013 at 17:42
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$\begingroup$ I had to type slightly less :) $\endgroup$ Commented Nov 5, 2013 at 17:44
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$\begingroup$ Thanks a lot Jan! Is it possible to determine $spec(S)$? In some cases, it looks like $G\cdot X$ but I'm not sure if it's true in general. $\endgroup$– NN guestCommented Nov 5, 2013 at 20:37
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$\begingroup$ What do you mean by $G \cdot X$? $\endgroup$ Commented Nov 6, 2013 at 9:54
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$\begingroup$ I'm sorry for not being clear in my last question. I just didn't understand what "affinization" means. I was considering the example when $X=\mathfrak{u}$, the Lie algebra of the unipotent radical subgroup of $B$, and $B$ acts on it as conjugation, denoted by a dot "$\cdot$". Then we have $G\cdot\mathfrak{u}$ is the nilpotent cone of Lie$(G)$ which is the spectrum of $(k[G]\otimes k[\mathfrak{u}])^B$. $\endgroup$– NN guestCommented Nov 6, 2013 at 10:48