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(The question of the type "how to define?")

Let $(R,\mathfrak{m})$ be a local ring over a field $k$ of zero characteristic. Consider the matrices over this ring, $Mat(m,R)$. I think of $Mat(m,R)$ as an infinite dimensional affine $k$-scheme.

Consider the group over this ring, $GL(m,R)$, e.g. acting from the left. Take some "good enough" subgroup $G\subseteq GL(n,R)$. For a fixed matrix $A$ consider the orbit $GA\subset Mat(m,R)$. I'd like to think of $GA$ as some subscheme of $Mat(m,R)$. At least to consider the germ of the orbit at the given point $(GA,A)$ as a subspace of the germ $(Mat(m,R),A)$. (Even though it is usually of infinite dimension and infinite codimension.) I'd like to have the notion of Zariski tangent space $T_{(GA,A)}$, which is a $k$-subspace of $T_{(Mat(m,R),A)}$.

If $R$ is $\mathfrak{m}$-complete then one can consider the "jet" projections, $R\stackrel{jet_p}{\rightarrow}R/\mathfrak{m}^{p+1}$. Then one asks $jet_p(GA)$ to be a $k$-scheme. And $jet_p(GA,A)$ to be the germ of this scheme. (In this sense $G$ is "good enough"). And then defines everything as the projective limit. In particular, the tangent space can be computed via $\mathfrak{m}/\mathfrak{m}^2$ or via the maps of $Spec(k[\epsilon]/\epsilon^2)$ etc.

Probably one can do similarly if $R$ embeds into its $\mathfrak{m}$-completion.

But what is the general way to define? (e.g. if $R$ is the ring of germs of smooth functions then the orbit $GA$ and its completion can vary drastically.) This is surely written somewhere?

(One could take just some Hamel basis of $R$ over $k$, but then everything will be of uncountable dimension/codimension. Probably this is not a good way?)

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  • $\begingroup$ You can always use Weil restriction to get a tangent space object. Do you intend to impose any finiteness condition on $R/\mathfrak{m}$ as an extension of $k$? $\endgroup$
    – S. Carnahan ♦
    Commented Sep 11, 2014 at 3:49
  • $\begingroup$ Sorry, I'm confused and ignorant. Do you mean the restriction $R\rightarrow k$? In simple words, do you mean to impose that $GA$ is a scheme over $R$ and then to consider it over $k$? (And similarly to take the tangent space over $R$, then consider it over $k$?). $\endgroup$
    – Dmitry Kerner
    Commented Sep 11, 2014 at 9:57
  • $\begingroup$ >>finiteness condition on R/m as an extension of $k$? $R$ is a local ring over the field $k$, so $R/\mathfrak{m}=k$. Or did you mean smth else? $\endgroup$
    – Dmitry Kerner
    Commented Sep 11, 2014 at 9:58

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