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Let $G$ be a (complex algebraic) group, $H$ a subgroup, and ${\cal O}(G)$ and ${\cal O}(H)$ the coordinate algebras of complex regular functions of $G$ and $H$ respectively. Can ${\cal O}(H)$ ever embedded as a subalgebra of ${\cal O}(G)$? For example, I'm looking at $U(k-1)$ as a subalgebra of $SU(k)$ (embedded in the bottom right-hand corner with $1$ in the $u^1_1$ entry and $0$ everywhere else) and can't an embedding of ${\cal O}(U(k-1)$) into ${\cal O}(SU(k)$).

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up vote 1 down vote accepted

It may be possible in rare circumstances, but the natural thing is that $\mathcal{O}(H)$ is a quotient of $\mathcal{O}(G)$, not a subalgebra. The quotient map is dual to the inclusion $H\to G$.

I guess in the case that $G$ is a direct product $H \times K$, you get $$ \mathcal{O}(G) \simeq \mathcal{O}(H) \otimes \mathcal{O}(K), $$ and in this case you have what you want. But in general, the arrow should go the other direction.

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