22
$\begingroup$

This question is to add to my understanding why the concept of a determinant does not extend to an infinite dimensional vector space. I am already aware of a couple facts which hint why this is so:

  • The determinant of an endomorphism of a finite dimensional vector space with dimension $n$ can be defined in a basis-free way as the composition of these canonical maps: $$\mathrm{End}(V)\xrightarrow{\phi} \mathrm{End}(\Lambda^n V)\xrightarrow{\psi} K$$ defined by $\phi(A)=((x_1\wedge\cdots\wedge x_n)\mapsto(Ax_1\wedge\cdots\wedge Ax_n))$ and $\psi$ defined as the inverse of the map $\psi^{-1}:K\rightarrow \mathrm{End}(\Lambda^n V)$ defined by $\psi^{-1}(\lambda)=(x\mapsto \lambda x)$. This construction reveals why finite dimension is important: $\mathrm{End}(\Lambda^n V)$ need not be 1-dimensional otherwise for any $n$ if $V$ fails to be finite dimensional. And thus our last map fails to exist.
  • Another reason that the determinant fails to extend to infinite dimensional spaces is that there are injective linear endomorphisms which do not have an inverse. Such maps may still have left inverses, but no right inverse. Such a pair is the right-shift and left-shift maps $$(x_1,x_2,\ldots)\mapsto(0,x_1,x_2,\ldots)\qquad (x_1,x_2,\ldots)\mapsto(x_2,x_3,\ldots)$$ where the left-shift is the left inverse of the right-shift; however, the left-shift remains non-invertible. A 'good' generalization of determinant would assign non-zero determinant to the first and zero to the last. This results in the determinant of the inverse not being the inverse of the determinant.
  • Another way to see that the concept does not generalize is that if a determinant for an operator exists, you might expect it to be the product of the eigenvalues. In general, a linear endomorphism from an infinite dimensional space can have infinitely many eigenvalues.

The previous fact suggests that we could define the determinant for a specific subset of $\mathrm{Aut}(V)$, namely those automorphisms which fix all but a finite number of 1-dimensional subspaces of $V$. But how far could we go with this generalization? Once you show that the determinant exists for a finite dimensional vector space, you can interpret the determinant as a nontrivial map which restricts to a group homomorphism from $\mathrm{Aut}(V)$ to $K^\times$ and assigns $0$ to the rest of the endomorphisms. Can we show that there is no nontrivial homomorphism from $\mathrm{Aut}(V)$ to $K^\times$ for an infinite-dimensional vector space? Much like we can show that the sign homomorphism does not extend to $S_{\Bbb N}$?

$\endgroup$
4
  • 1
    $\begingroup$ Actually there are:en.wikipedia.org/wiki/Functional_determinant $\endgroup$ – Peter Wu Feb 21 '14 at 2:22
  • $\begingroup$ That's extremely interesting. But it seems like the definition indirectly relies on a norm for the vector space to extend the definition of a trace. $\endgroup$ – user123641 Feb 21 '14 at 2:32
  • $\begingroup$ @Bryan: I don't quite understand the uniqueness statement in your last paragraph. If $V$ is a finite dimensional vector space, then any multiplicative function $f : \text{End}(V) \to K$, which satisfies $f(0) \ne f(1)$, must be nonzero precisely on $\text{Aut}(V)$. Thus there is a bijection $\text{Hom}_{\text{Grp}}(\text{Aut}(V), K^\times) \leftrightarrow \{f : \text{End}(V) \to K \mid f \text{ multiplicative}, f(0) \ne f(1)\}$, and there are many such maps, not just $\det$ $\endgroup$ – zcn Feb 26 '14 at 7:04
  • $\begingroup$ You're right. I don't know what made me think that. Edited $\endgroup$ – user123641 Feb 27 '14 at 6:24
4
$\begingroup$

Since you already seem satisfied with the argument that the sign homomorphism does not extend to the group of (not necessarily finitely supported) permutations of $\Bbb N$, consider that by acting on the canonical basis of $K[X]$, such permutations define an isomorphic subgroup of automorphisms of this infinite dimensional vector space, and that an extension of the determinant to $\operatorname{Aut}_K(K[X])$ would by restriction to that subgroup define such an impossible extension of the sign homomorphism to permutations of $\Bbb N$.

$\endgroup$
5
  • $\begingroup$ Well it convinces me for any subfield of $\Bbb R$ and $\Bbb F_3$, but I do not see why it generalizes to any field $K$ (barring $\Bbb F_2$ of course). Is there always a homomorphism from $K^\times$ to $\{-1,1\}$? $\endgroup$ – user123641 Feb 24 '14 at 17:44
  • $\begingroup$ Well there is not always a surjective group homomorphism $K^\times\to\{-1,1\}$, for instance it appears there is none for $K=\Bbb C$ (I think I saw this mentioned in a recent question); there is one though for all finite fields (outside characteristic $2$). But I don't see how that enters into my answer: I'm just restriction a hypothetical determinant morphism to essentially set of infinite permutation matrices. You don't need any morphism on $K^\times$ to define the determinant of a permutation matrix. $\endgroup$ – Marc van Leeuwen Feb 24 '14 at 18:16
  • $\begingroup$ I think you're assuming that my candidate map of determinant would restrict to the sign homomorphism when considering only these automorphisms. Since the sign homomorphism doesn't extend to these automorphisms, there can be no such determinant map. But I'm still worrying about the case where the restriction need not reflect the sign homomorphism. $S_{\Bbb N}$ also has no surjective homomorphism onto $Z_2$, but that does not immediately bar the existence of a nontrivial homomorphism from $\mathrm{Aut}(V)$ to $K^\times$. We have however ruled out subfields of $\Bbb R$ and all finite fields. $\endgroup$ – user123641 Feb 25 '14 at 0:49
  • $\begingroup$ Since $S_{\Bbb N}$ has no nontrivial homomorphism to $Z_2$, any such determinant that was non-trivial would have to send $S_{\Bbb N}$ to 1. I agree this map might not look like a nice generalization already, but I want to know if all homomorphisms from $V$ to $K^\times$ are trivial. $\endgroup$ – user123641 Feb 25 '14 at 1:04
  • $\begingroup$ Nice argument. I should add that these facts (the sign of infinite permutations can't be defined) and the fact that we cannot extend determinant to infinite dimension in full generality, were first proved by Vitali in a 1915 note (G. Vitali. Sostituzioni sopra una infinità numerabile di elementi. Bollettino Mathesis 7: 29-31, 1915.). $\endgroup$ – YCor Jul 15 '19 at 14:51
9
$\begingroup$

My suspicion is that there are issues with solely considering purely algebraic infinite-dimensional vector spaces and trying to generalize the determinant from its algebraic construction.

However, there are analytic generalizations of determinants. Some of them are very deep and I am no expert on them, but there is at least one very concrete generalization I am familiar with.

Definition: Let $A$ be a finite rank operator on a Hilbert space. Then define $$\det(I - A) = \prod_j (1 - \lambda_j(A)),$$ where $\lambda_j$ is the $j$th largest eigenvalue of $A$. If $A$ is instead a trace class operator on a Hilbert space, then define $$\det(I - A) = \lim_{k \rightarrow \infty} \det(I - A_k),$$ where $\{A_k\}$ is a sequence of finite rank operators such that $\|A - A_k\|_{L^1} \rightarrow 0$.

$\endgroup$
2
  • 1
    $\begingroup$ For the sake of the OP's understanding, I would like to add that the Shift Operator is not trace class. Any trace class operator (or more generally any compact operator) is approximable in the operator norm by finite rank operators. The shift operator is not. $\endgroup$ – Joel Feb 24 '14 at 15:04
  • 2
    $\begingroup$ Is there any reference in the literature for the above definition? $\endgroup$ – La Rias Jun 2 '18 at 6:14
5
$\begingroup$

The answer to your question depends exactly on what you want the determinant to have.

A reasonable set of properties to ask on a determinant is that $$ \tag{0} \det\mathrm{id}_V = 1 $$ for all identity maps $\mathrm{id}_V:V\to V$ of all vector spaces $V$, that $$ \tag{1} \det(f\circ g)=\det f\cdot\deg g $$ whenever $f:V\to V$ and $g:V\to V$ are linear maps, and that $$ \tag{2} \det\begin{pmatrix}f&h\\0&g\end{pmatrix}=\det f\cdot\det g $$ whenever $f:V\to V$, $g:W\to W$ and $g:W\to V$ are linear maps (the matrix on the left denotes the hopefully obvious linear map $V\oplus W\to V\oplus W$).

Suppose we do have such a map and let $f:V\to V$ be an automorphism of a vector space $V$. Let $V^\infty=V\oplus V\oplus V\oplus\cdots$ be a countable sum of copies of $V$ and $f^\infty=f\oplus f\oplus f\oplus\cdots$ the obvious automorphism of $V^\infty$. Using only the above two properties of the function~$\det$ you can check that $\det f^\infty$ is a non-zero element of the field and that $$\det f\cdot\det f^\infty=\det f^\infty,$$ so that in fact $$\det f=1.$$

The conclusion of this is that if you want the define the determinant of automorphisms of vector spaces in such a way that (0), (1) and (2) hold, then you only can define it to be always $1$, not the most interesting definition.

$\endgroup$
1
  • 1
    $\begingroup$ If we look for functions $\det$ satisfying (0), (1) and (2) but restricted to finite dimensional vector spaces, we find that they are exactly those obtained by composing the usual determinat with a group homomorphism $\mathbb{K}^\times\mathbb{K}^\times$. All this is really a calculation of the first K-theory group of a field and categories; see the introduction to K-theory by Jonathan Rosenberg. $\endgroup$ – Mariano Suárez-Álvarez Jun 7 '20 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy