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Does every invertible linear map $M$ between $V \oplus V$ and $W \oplus W$ naturally yield an invertible linear map $L$ between $V$ and $W$?

Here "naturally" means "in an $GL(V) \times GL(W)$-equivariant fashion" [originally I wrote $SL(V)$ here and below, but Todd Trimble and Eric Wofsey pointed out my mistakes]. Cf. the article "Producing New Bijections From Old" by David Feldman and myself, available at http://jamespropp.org/cancel.pdf, in which we raise such questions in the context of maps between finite sets $S$ and $T$ and ask for $Sym(S) \times Sym(T)$-equivariance.

One might try to construct $L$ from $M$ by mapping $V$ to $V \oplus V$ (sending $v$ to $(v,0)$, say), mapping $V \oplus V$ to $W \oplus W$ (by $M$), and then mapping $W \oplus W$ to $W$ (sending $(w,w')$ to $w$, say). But the map $L$ constructed in this fashion need not be invertible; e.g., if $V = W$ and $M$ sends $(v,v')$ to $(v',v)$, then $L$ is the 0-map.

Something cleverer is called for; or maybe by paying attention to $GL(V)$ one can show that no such general construction is possible.

If there is an existing literature on problems of this kind, pointers would be appreciated.

(Note that Conway, Doyle, and Qiu, in http://arxiv.org/abs/math/0605779 and http://arxiv.org/abs/1504.01402, take a slightly different approach to questions of this kind. I'm not sure what a "linear" version of their approach would be.)

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    $\begingroup$ Note that one test of naturality in this case is whether it is true for infinite-dimensional vector spaces without the axiom of choice. $\endgroup$ – Greg Kuperberg Apr 10 '15 at 1:55
  • $\begingroup$ Why $SL(V)$ rather than $GL(V)$? In the latter case, the problem seems to be whether there is a map $\hom(2 \cdot V, 2 \cdot W) \to \hom(V, W)$ that is natural in the arguments $V, W$ which are objects of the groupoid of (possibly finite-dimensional) vector spaces and linear isomorphisms. $\endgroup$ – Todd Trimble Apr 10 '15 at 2:31
  • $\begingroup$ You're quite right! I wrote "$SL$" unthinkingly. I'll fix the post (leaving tracks so that your comment still makes sense to later readers). $\endgroup$ – James Propp Apr 10 '15 at 3:01
  • $\begingroup$ Shouldn't $SL(V)$-equivariant maps also be $GL(V)$-equivariant at least in characteristic $0$ ? $\endgroup$ – darij grinberg Apr 10 '15 at 3:17
  • $\begingroup$ I assume you mean to ask for equivariance with respect to $GL(V)\times GL(W)$, not just with respect to $GL(V)$ (since $GL(V)$ alone acts freely on both sides). If you identify $V$ and $W$, then the stabilizer of a point in $Iso(V,V)$ is conjugate to the diagonal in $GL(V)\times GL(V)$, so the question is: given an element $M\in Iso(V^2,V^2)$, is the stabilizer of the action of $GL(V)\times GL(V)$ at $M$ contained in a conjugate of the diagonal? $\endgroup$ – Eric Wofsey Apr 10 '15 at 3:18
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For infinite-dimensional vector spaces this not possible. According to this paper of Brenner and Ringel, if $A$ is any principal ideal domain that is not a field or a complete valuation ring, then there is an $A$-module $V$ such that $V\cong V\oplus V\oplus V$ but $V\not\cong V\oplus V$. In particular, if $k$ is a field and $A=k[\mathbb{Z}]=k[x,x^{-1}]$, let $V$ be such a module and $W=V\oplus V$. Then $V$ and $W$ are representations of $\mathbb{Z}$ such that $V\oplus V\cong W\oplus W$ but $V\not\cong W$. As explained in Will's answer, this gives a counterexample to the original question. Explicitly, let $M:V\oplus V\to W\oplus W$ be any $k$-linear isomorphism that is also $A$-linear. The action of $x$ on $V$ and $W$ then gives an element $x\in GL(V)\times GL(W)$ which stabilizes $M$. However, since $V$ and $W$ are not isomorphic as $A$-modules, there is no isomorphism $L:V\to W$ which is stabilized by $x$. Thus there can be no $GL(V)\times GL(W)$-equivariant assignment of an isomorphism $L:V\to W$ corresponding to $M$.

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Division by two, with your naturality condition, is equivalent to division by two in the category of representations of a group:

If $V$ and $W$ are two finite-dimensional representations of a group $G$ such that $V + V = W+ W$, then $V=W$.

As Eric said, this is all about stabilizers. Given no other conditions, there is a $GL(V) \times GL(W)$-equivariant map from one set to another if the stabilizers of the action on the first set are contained in the action on the other.

So fix an isomorphism $V + V = W +W $ and let $G$ be its stabilizer. This makes $V$ and $W$ into $G$-representations, where $V+V$ and $W+W$ are isomorphic $G$-representations. Then $G$ is conjugate to a diagonal subgroup if and only if $V$ and $W$ are isomorphic as representations.

Unfortunately it's not obvious to me if division by two in the category of representations is true. It's clearly true if we take semisimple representations, by unique decomposition into irreducibles, but is the decomposition into indecomposables unique?

Edit: As Greg-Kuperburg says, the answer is "yes" by the Krull-Schmidt theorem. So the answer is yes, but this does not give an explicit construction.

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    $\begingroup$ Wikipedia says that this assertion, the "Krull-Schmidt theorem", holds for finite-length modules over a ring, but not in general for modules that are only Noetherian or only Artinian. en.wikipedia.org/wiki/… $\endgroup$ – Greg Kuperberg Apr 10 '15 at 4:24
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    $\begingroup$ On the other hand, this conference proceedings, "Infinite Length Modules" by Krause and Ringel, says that Krull-Schmidt can fail for infinite-dimensional representations of a finite group, over an infinite field with positive characteristic. books.google.com/… $\endgroup$ – Greg Kuperberg Apr 10 '15 at 4:38
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    $\begingroup$ @GregKuperberg I wonder if this can be use to construct a counterexample to the infinite-dimensional version of this problem. $\endgroup$ – Will Sawin Apr 10 '15 at 4:40
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    $\begingroup$ Yeah, it looks like it just is a counterexample. $\endgroup$ – Greg Kuperberg Apr 10 '15 at 4:40
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    $\begingroup$ @GregKuperberg in characteristic zero or just in characteristic p? $\endgroup$ – Will Sawin Apr 10 '15 at 4:43
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Here is a sketch of a more explicit answer than the one referred to by Eric Wofsey. (The paper by Brenner and Ringel is quite abstract, and obtains results that are both harder and more general.)

In the paper On direct decompositions of torsionfree abelian groups, Jónsson constructs the following failure of division by two in the category of abelian groups: Let $P$ and $Q$ be two infinite, disjoint sets of primes other than 5, and let $A$ be the subgroup of $\mathbb{Q}^2$ generated by all all $(1/p,0)$ with $p \in P$, all $(0,1/q)$ with $q \in Q$, and the point $(1/5,1/5)$. Then $A$ is indecomposable and fairly rigid. We get that $A \otimes \mathbb{Q}^2 \cong \mathbb{Q}^2$, then that any element of $\text{Aut}(A) \subseteq \text{SL}(2,\mathbb{Q})$ has a diagonal matrix, then that it has to be scalar multiplication by $\pm 1$. Likewise let $B$ be generated by all $(1/p,0)$, $(0,1/q)$, and $(1/5,2/5)$. Then the same sort of rigidity shows that $A$ and $B$ aren't isomorphic. They both tensor to the same $\mathbb{Q}^2$, but any element of $\text{SL}(2,\mathbb{Q})$ that sends one to the other would have to be diagonal and then cannot exist.

On the other hand, there is an explicit linear transformation in $\text{SL}(4,\mathbb{Q})$, given by applying an element of $\text{SL}(2,\mathbb{Z})$ to the first and third coordinates and another to the second and fourth coordinates, that takes $A \oplus A$ to $B \oplus B$.

I haven't checked the details, but I would expect this to work for any PID with infinitely many inequivalent primes, in particular as Eric says for the PID $k[x,x^{−1}]$, where $k$ is any field. Then you get two infinite-dimensional vector spaces $A$ and $B$, with non-isomorphic indecomposable linear representations of the infinite cyclic group, and with the same property $A \oplus A \cong B \oplus B$.

I wonder whether the answer to Jim's question would be different for Hilbert spaces considered up to isometry. By the spectral theorem combined with the Hahn-Hellinger theorem, $A \oplus A \cong B \oplus B$ implies $A \cong B$ at least for separable Hilbert spaces which are representations of a finitely generated abelian group. (Maybe also a countably generated abelian group, I'm not sure.)

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This comment is about the "natural" part of the question. Suppose we move to Banach spaces so that one naturally expects that if one has a bounded linear isomorphism between $V\oplus V$ and $W\oplus W$, then the corresponding linear isomorphism between $V$ and $W$ should be also bounded. Well, in this case the answer is negative: Gowers and Maurey constructed in "Banach spaces with small spaces of operators (1997)" examples of Banach spaces $V$ which are isomorphic to their "cubes" $V\oplus V \oplus V$ but not to the square $V\oplus V$. Therefore if we set $W=V\oplus V$, then we have that $V\oplus V$ is isomorphic to $W\oplus W$, while $V$ and $W$ are not isomorphic (as Banach spaces, I mean).

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