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In a recent thread I show that if $V$ is any infinite-dimensional vector space, then (assuming the axiom of choice) the monoid $\text{End}(V)$ has trivial abelianization. It follows in particular that any multiplicative function

$$\det : \text{End}(V) \to K$$

where $K$ is the underlying field (for me "multiplicative" includes the unit condition $\det(1) = 1$) must be identically equal to $1$.

In this thread from 2020 Mariano shows that if we ask for a determinant $\det : \text{Aut}(V) \to K$ on automorphisms of infinite-dimensional vector spaces which is both multiplicative and compatible with direct sums of vector spaces, then again $\det(f) = 1$ identically.

The arguments used in these threads don't appear to generalize to the following:

Q1: What if we only ask for a determinant $\det : \text{Aut}(V) \to K^{\times}$ on a single infinite-dimensional vector space $V$, and ask only for multiplicativity?

That is, can we show that if $\det : \text{Aut}(V) \to K^{\times}$ satisfies $\det(1) = 1$ and $\det(fg) = \det(f) \det(g)$, then $\det(f) = 1$ identically?

In this case I believe that more is true: namely, I believe that if $V$ is infinite-dimensional then $\text{Aut}(V)$ has trivial abelianization (so the choice to map into $K^{\times}$ specifically shouldn't matter) and is even "almost simple." More precisely, assuming that $V$ is countable-dimensional, I believe $\text{Aut}(V)$ has a unique maximal normal subgroup given by automorphisms which are a finite-rank perturbation of the identity $I + K$, or equivalently which fix a cofinite (finite-dimensional quotient) subspace of $V$, and that the quotient by this normal subgroup is simple.

Q2: How do we prove these two assertions, if they're true?

The corresponding fact about sets is that $\text{Aut}(\mathbb{N})$ (the "big" infinite symmetric group) has a unique maximal normal subgroup $S_{\infty}$ (the "small" infinite symmetric group) given by permutations which fix all but finitely many points, and that the quotient by this normal subgroup is simple, which I believe I read somewhere but also don't know how to prove.

Qiaochu Yuan
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    If $V$ is an infinite-dimensional vector space over a field, then $[Aut(V),Aut(V)] = Aut(V)$. See Corollary 3.5 (and the remark after it) in "A. Rosenberg, The structure of the infinite general linear group. Ann. of Math. (2) 68 (1958), 278–294." https://doi.org/10.2307/1970248 – Mikko Korhonen Mar 17 '25 at 05:28
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    @MikkoKorhonen: I think this should be an answer, not a comment. – Moishe Kohan Mar 17 '25 at 19:50

1 Answers1

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Briefly summarizing Mikko Korhonen's comment: both questions are answered in Rosenberg's The structure of the infinite general linear group, which classifies the normal subgroups of $\text{Aut}(V)$. The proof technique is to consider elements of the form $1 + N$ where $N^2 = 0$ but the details of the arguments are unfortunately somewhat involved, more than can be easily summarized here, as far as I can tell.

Q1 is answered by Corollary 3.5 and Q2 is answered by Theorem B.

Qiaochu Yuan
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