Let $\Gamma$ be a nontrivial subgroup of $GL(n,\mathbb{C})$. Then does the linear span of $\Gamma$ over $C$ have to contain the algebraic closure of $\Gamma$? Or is there some required condition in order that this statement to be true?
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1$\begingroup$ By linear span you mean the subspace spanned by $\Gamma$ in the vector space of matrices? By algebraic closure you mean the Zariski closure? $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 5, 2011 at 18:11
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1$\begingroup$ Let $V$ be a finitedimensional $C$-vector space and $S \subset V$ a set. Let $x$ be in the Zariski closure of $S$. By definition, each polynomial that vanishes on $S$ has to vanish at $x$. Take a linear form $f$ that vanishes on $S$. Since $f$ is a polynomial, it follows $f(x)=0$. Thus any linear form which is trivial on $S$ is trivial on the Zariski closure. Conclusion: the Zariski closure is a subset of the linear span. So the answer is yes, and has nothing to do with groups. $\endgroup$– Johannes EbertCommented Mar 5, 2011 at 18:28
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2$\begingroup$ If Mariano has guessed the definitions correctly, then the "linear span" of $\Gamma$ is closed and contains $\Gamma$, hence contains the Zariski closure of $\Gamma$. The fact that $\Gamma$ is a group has nothing to do with it. $\endgroup$– Steven LandsburgCommented Mar 5, 2011 at 18:33
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$\begingroup$ Looks like Johannes beat me to this. $\endgroup$– Steven LandsburgCommented Mar 5, 2011 at 18:33
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$\begingroup$ But only by a second. $\endgroup$– Johannes EbertCommented Mar 5, 2011 at 19:31
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Furthermore the linear span of $\Gamma$ is not contained in $GL(n,\mathbb{C})$. So you have to ask, where is it Zariski closed?
