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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMay 22nd 2017
   • (edited May 22nd 2017)
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   Author: Urs
   Format: MarkdownItexat _[[general linear group]]_ we only had some sentences on its incarnation as an algebraic group. I have started a subsection _[Definition -- As a topological group](https://ncatlab.org/nlab/show/general+linear+group#AsAtopologicalGroup)_ with some basics.

   at general linear group we only had some sentences on its incarnation as an algebraic group. I have started a subsection Definition – As a topological group with some basics.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 29th 2017
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   Author: Urs
   Format: MarkdownItexWhat's a quick proof that the topology on $GL(n,k)$ as a subspace of $\mathbb{R}^{n^2}$ with its Euclidean topology coincides with that as a subspace of $Maps(k^n,k^n)$ with its compact-open topology?

   What’s a quick proof that the topology on $GL(n,k)$ as a subspace of $\mathbb{R}^{n^2}$ with its Euclidean topology coincides with that as a subspace of $Maps(k^n,k^n)$ with its compact-open topology?

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 29th 2017
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   Author: Todd_Trimble
   Format: MarkdownItexI guess $k = \mathbb{R}$? So we do have an injective continuous map $GL(n, \mathbb{R}) \to Map(\mathbb{R}^n, \mathbb{R}^n)$, the currying of the continuous action $GL(n, \mathbb{R}) \times \mathbb{R}^n \to \mathbb{R}^n$. This says that the usual Euclidean topology is finer than the subspace topology coming from $Map(\mathbb{R}^n, \mathbb{R}^n)$. But the Euclidean topology is also coarser. Let's do this for $M(n, \mathbb{R})$ instead of $GL(n, \mathbb{R})$. A Euclidean neighborhood base of a linear map or matrix $A$ consists of sets of the form $\{B: \forall_{1 \leq i \leq n}\; |A e_i - B e_i| \lt \epsilon\}$. But this is a basis/base element for the function space topology, $\bigcap_{i = 1}^n C(K_i, U_i)$ where $K_i = \{e_i\}$ and $U_i$ is the $\epsilon$-ball about $A e_i$. (It may help to think of convergence in the function space topology as the exact same as uniform convergence over every compact set.)

   I guess $k = \mathbb{R}$?

   So we do have an injective continuous map $GL(n, \mathbb{R}) \to Map(\mathbb{R}^n, \mathbb{R}^n)$, the currying of the continuous action $GL(n, \mathbb{R}) \times \mathbb{R}^n \to \mathbb{R}^n$. This says that the usual Euclidean topology is finer than the subspace topology coming from $Map(\mathbb{R}^n, \mathbb{R}^n)$.

   But the Euclidean topology is also coarser. Let’s do this for $M(n, \mathbb{R})$ instead of $GL(n, \mathbb{R})$. A Euclidean neighborhood base of a linear map or matrix $A$ consists of sets of the form $\{B: \forall_{1 \leq i \leq n}\; |A e_i - B e_i| \lt \epsilon\}$. But this is a basis/base element for the function space topology, $\bigcap_{i = 1}^n C(K_i, U_i)$ where $K_i = \{e_i\}$ and $U_i$ is the $\epsilon$-ball about $A e_i$.

   (It may help to think of convergence in the function space topology as the exact same as uniform convergence over every compact set.)

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMay 29th 2017
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   Author: Urs
   Format: MarkdownItexThanks, Todd! I have added that to the entry [here](https://ncatlab.org/nlab/show/general+linear+group#AsSubspaceOfTheMappingSpace).

   Thanks, Todd!

   I have added that to the entry here.