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Let $V$ be a topological vector space (not necessarily finite-dimensional) over a field $K$, and let $GL(V)$ be the group of invertible linear maps $V\to V$ under composition. There are two obvious topologies for $GL(V)$:

  1. The compact-open topology, which is generated by a subbase $\{f\in GL(V):f[C]\subseteq U\}$ for $C$ compact, $U$ open;

  2. Choose a basis $B$ for $V$, and interpret $GL(V)$ as elements of $K^{B\times B}$; topologize this set by the product topology.

My question is: Do these topologies coincide, and is $GL(V)$ a topological group under these topologies? I think I can show that (2) is independent of the basis:

Suppose that $B$ and $B'$ are bases for $V$; to show that the topology (2) with respect to $B$ is the same as the one using $B'$, it is sufficient to show that the subbase for $B'$ is open with respect to the $B$-topology. A subbase element in the $B'$-topology is a translate of a set of the form $\{f:f(b)_a\in U\}$ for some $a,b\in B'$ and $0\in U\subseteq K$ open (where $x_a$ is the $a$-th component of the $B'$ basis representation for $x$, $x=\sum_{a\in B'}x_aa$). Since $B$ is a basis, $a$ has a $B$ basis representation $a=\sum_{v\in B}a_vv$, and since $V$ is a TVS for fixed $n$ there is a neighborhood $U'\subseteq K$ of zero such that all $n$-element sums from $U'$ are in $U$. Then since $f(b)_a=\sum_{\alpha\in B}\sum_{\beta\in B}a_\alpha b_\beta f(\beta)_{\alpha}$ is a finite sum, choosing $n\ge|\{\alpha\in B:a_\alpha\ne0\}||\{\beta\in B:b_\beta\ne0\}|$ we have

$$\bigcap_{\alpha\in B:a_\alpha\ne0}\bigcap_{\beta\in B:b_\beta\ne0}\{f:a_\alpha b_\beta f(\beta)_{\alpha}\in U'\}\subseteq \{f:f(b)_a\in U\}.$$

Does either of (1) or (2) hold the claim of being the "natural" topology for $GL(V)$? I saw a topology textbook discuss (2) in the context of finite-dimensional matrices, but I don't know if there are undesirable properties of one definition or the other in the infinite-dimensional setting.

Mario Carneiro
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    Usually when dealing with the case of a topological vector space (or a Hilbert space or a normed linear space), we are really only concerned with the elements of $GL(V)$ which are continuous with continuous inverse. If $V$ has more structure then a topology there might be different suitable topologies to choose from. For example if $V$ is a normed space, one might want to look at $V$ thought of as a subset of the space of continuous linear operators with the operator norm topology or with the weak topology with respect to that norm. IIRC the latter is equivalent to the compact open topology. – PVAL-inactive Jul 07 '15 at 17:42
  • @PVAL Oh yes, I forgot to mention how all of these questions are affected by restricting to linear homeomorphisms of $V$. Indeed usually the compact open topology is only used for spaces of continuous functions (although the definition makes sense even for discontinuous functions). Do the two topologies mentioned coincide on this subgroup $CL(V)$? I don't think there are any other natural options if we only have a TVS, with no norm. – Mario Carneiro Jul 07 '15 at 17:56
  • This question is very similar to http://math.stackexchange.com/q/37528/50776 but I would like to focus more on the infinite dimensional case, as well as the question of how (or if) the different topologies are different from each other in common cases. – Mario Carneiro Jul 07 '15 at 18:10

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