Let $K$ be a field and $G$ be an algebraic group. Specifically $O(n)$ or $Sp_{2n}$. Is it true that for any ring $A$ over $K$ , $G(A)\cong G(A[x])$. Is there any reference for such kind of results?
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3$\begingroup$ Certainly not with $G=\Bbb{G}_a$. $\endgroup$– abxCommented Aug 15, 2018 at 10:17
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2$\begingroup$ Yes, For G=$GL_n$ also it is not true. Thats why I asked for $O(n)$ and $Sp_{2n}$. $\endgroup$– GirishCommented Aug 15, 2018 at 10:19
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3$\begingroup$ It is not true for $\operatorname{SL}_2 $, hence not for $\operatorname{Sp}_{2n} $. $\endgroup$– abxCommented Aug 15, 2018 at 10:56
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2$\begingroup$ @abx If $K$ is of infinite dimension over it prime subfield, call it $F$, then $\mathbb{G}_a(A)$ and $\mathbb{G}_a(A[x])$ are both $F$-vector spaces of $F$-dimension $\dim_FA=\aleph_0\cdot \dim_F A$ and are therefore isomorphic as abstract groups for any $K$-algebra $A$. $\endgroup$– Uriya FirstCommented Aug 15, 2018 at 11:11
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2$\begingroup$ @Girish In that case, as soon as $G$ contains a copy of $\mathbb{G}_a$ as a closed subgroup, the map $G(A)\to G(A[x])$ will not be an isomorphism. This is always the case when $K$ is algebraically closed and $G=\mathrm{O}_n$ with $n>2$ or $\mathrm{Sp}_n$ with $n>1$ (the unipotent radical of a Borel subgroup will contain such a copy). Since you can always take $A$ to an algebraically closed field, $G(A)\to G(A[x])$ will not be an isomorphism for all $A$-s when $G=\mathrm{O}_n$ or $\mathrm{Sp}_n$. $\endgroup$– Uriya FirstCommented Aug 15, 2018 at 15:06
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1 Answer
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The comments show that the answer to the question is negative if $G$ contains a copy of $\mathbb G_a$. The examples $O(n)$ and $Sp(n)$ in the question suggest, though, that the emphasis is on anisotropic $G$. It is interesting that in this case the answer is affirmative to some extent.
Let, e.g., $K=\mathbb R$, let $G$ be a linear algebraic group not containing $\mathbb G_a$. This means that $G^0=SC$ where $S$ is a split torus and $C$ is anisotropic, i.e., $C(\mathbb R)$ is compact. Claim: Under these conditions $G(K)\overset\sim\to G(K[x])$.
An element of $G(K[x])$ is a morphism $f:\mathbb A^1\to G$. If $G=S$ then $f$ is constant since $K[x]^\times=K^\times$. If $G=C$ then the image of $f$ is closed. The reason for this is that $C$ is affine and therefore $f$ can't extend to $\mathbb P^1$. But, since $C(\mathbb R)$ is compact, the image of $\mathbf A^1(\mathbb R)$ in $C(\mathbb R)$ can't be closed unless $f$ is constant. The mixed case follows from these two subcases.
From this it follows easily that $G(A)\overset\sim\to G(A[x])$ when $A=\mathcal O(X)$ where $X$ is an irreducible normal affine $K$-variety such that $X(K)$ is Zariski dense in $X$.
With some effort, the argument can be extended to any field of characteristic zero.
answered Aug 19, 2018 at 22:36