# Are closed group of $GL(n)$ always determined by a set of tensors?

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 $$G=G_{\mathcal{T}}$$?

• 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 $\mathbb{R}$. Aug 27, 2023 at 9:59

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