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Let $\mathcal{T}$ be a set of tensors in $\mathbb{R}^n$, and let $G_{\mathcal{T}}$ be a subgroup of $GL(n)$ defined as $$ G_{\mathcal{T}} = \{ g \in GL(n) \mid g \cdot T = T, \, \forall T \in \mathcal{T} \}, $$ where "$\cdot$" denotes the appropriate tensor transformation rule under the action of $g \in GL(n)$. I know $G_{\mathcal{T}}$ is a closed subgroup of $GL(n)$ (right?).

Now, given a closed subgroup $G$ of $GL(n)$, can we find a set of tensors $\mathcal{T}$ such that $G=G_{\mathcal{T}}$?

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    $\begingroup$ This is a polynomial condition, so such a subgroup will always be closed in the Zariski topology (not just the Euclidean topology). I convinced myself at some point that the converse ought to be true over $\mathbb{C}$ because of Tannaka-Krein but I'm not sure about $\mathbb{R}$. $\endgroup$ Aug 27, 2023 at 9:59

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No, let $G\neq GL_n(\mathbb{R})$ be a closed subgroup which contains $\lambda I_n$ for some $|\lambda|> 1$, e.g. $\mathbb{Z}^n$ or diagonal matrices. Suppose $G=G_{\mathcal{T}}$. Then $(\lambda I_n)\cdot T =\lambda^k T$ for a $k$-tensor $T$, and so, $(\lambda I_n)\cdot T=T$ implies $T=0$. Therefore, since $G\subseteq G_{\mathcal{T}}$ it is $\mathcal{T}\subseteq\{0\}$ and $G_{\mathcal{T}}=GL_n(\mathbb{R})\neq G$, a contradiction.

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  • $\begingroup$ Thanks! Do you know what under what conditions over $G$ the result is true? $\endgroup$ Aug 27, 2023 at 16:03

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