Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $C$ be the category of finite-dimensional vector spaces over some field. It is easy to construct pairs of endofunctors $F, G$ of $C$, of the same variance, such that $F(V)$ and $G(V)$ have the same dimension for every $V \in C$, yet are not naturally isomorphic.

For example, we could take $G(V) := (V^*)^{\otimes 2}$, and $F(V) := (V^{\otimes 2})^*$, where $*$ denotes the dual. Or, I'm sure many of us have, at some time, tried showing that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, before realizing that it is false. A convincing reason why they are not canonically isomorphic is that there is no obvious reason, in terms of universal properties, why they should be. In other words: if you try proving that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, you're going to have a bad time.

However, how does one explicitly show that $F$ and $G$ are not isomorphic? Is there a slick, systematic way of doing this?

Edit: Well, my examples couldn't be worse, as Mariano points out. But the gist of my question still stands (perhaps just get rid of the finite-dimensionality assumption ;))

share|improve this question

3 Answers 3

up vote 6 down vote accepted

Here is one way to construct families of endofunctors $F, G$ of $\text{FinVect}$ that can't be distinguished on the basis of dimension counts. First, consider the family of endofunctors $V \mapsto S^n(V)$, and second, consider the family of endofunctors $V \mapsto \Lambda^n(V)$. The first sends vector spaces of dimension $d$ to vector spaces of dimension ${d+n-1 \choose n}$ while the second sends vector spaces of dimension $d$ to vector spaces of dimension ${d \choose n}$. Both of these form bases for the space of integer-valued polynomials in one variable ($d$), hence together there are many linear dependences between them. The smallest one is that

$${d+1 \choose 2} = {d \choose 2} + d$$

which implies that $F(V) = S^2(V)$ and $G(V) = \Lambda^2(V) \oplus V$ can't be distinguished by their dimension counts.

Nevertheless, they are not isomorphic. The reason is that for any endofunctor $F$, the vector space $F(V)$ naturally admits the structure of a $\text{GL}(V)$-representation, and $S^2(V)$ and $\Lambda^2(V) \oplus V$ are not isomorphic as $\text{GL}(V)$-representations. More simply, you can consider the action of a diagonal matrix on $S^2(V)$ and on $\Lambda^2(V) \oplus V$ and verify that the traces (which are symmetric functions) are different.

In fact I think every endofunctor (edit: defining a polynomial function between Hom spaces, see Steve's comment) of $\text{FinVect}$ is equivalent to a direct sum of Schur functors, so they can all be distinguished in this way up to isomorphism. (I need some mild hypothesis on the base field; I think characteristic zero suffices.)

share|improve this answer
    
Nice example, +1. "In fact I think every endofunctor of FinVect is equivalent to a direct sum of Schur functors..." Probably you need to assume the functors are polynomial on Hom spaces (obviously direct sums of Schur functors are). –  S123 May 24 '13 at 23:11
    
Oh, hmm. I suppose there are endofunctors induced from endomorphisms of the base field and that these can be distinguished by applying the endofunctors to a one-dimensional vector space... –  Qiaochu Yuan May 24 '13 at 23:16
    
Good point, char. 0 is also necessary. –  S123 May 24 '13 at 23:37
    
Yes, I think something funny happens in positive characteristic because of divided powers, e.g. I think $S^p(V)^{\ast}$ and $S^p(V^{\ast})$ fail to be isomorphic in characteristic $p$, but I'm not sure about the details. –  Qiaochu Yuan May 25 '13 at 4:35
    
Just to be sure: By $S^n(V)$ you mean $\mathrm{Sym}^n(V)$, the $n$th symmetric power, right? –  Martin Brandenburg May 25 '13 at 10:00

The functors $FV=(V^*)^{\otimes 2}$ and $G(V)=(V^{\otimes2})^*$ on the category of finite fimensional vector spaces are isomorphic. The map $\phi:(V^*)^{\otimes2}\to(V^{\otimes2})^*$ such that $\phi(f\otimes g)(v\otimes w)=f(v)g(w)$ is a natural transformation (defined on the whole category of vector spaces) which in the category of f.d. vector spaces is an isomorphism of functors.

Similarly, $\Lambda(V^*)$ and $(\Lambda V)^*$ are canonically isomorphic on the category of f.d. vector spaces.

share|improve this answer
4  
One thing that does happen is that it is impossible to write down the inverse function for my map in a natural way: one does need to pick a basis and its dual basis to do so. But the resulting map ends up being independent of the choice of the basis, so nothing breaks. –  Mariano Suárez-Alvarez May 24 '13 at 21:41
1  
Also, $\phi$ is not an isomorphism if $V$ is infinite dimensional. –  Mariano Suárez-Alvarez May 24 '13 at 21:42
    
+1, You are absolutely right! I think that the origin of my misconception, regarding these two examples, is that there does not seem to be a way to appeal directly to the universal property in order to construct the isomorphism (or else it would hold regardless of dimension). It seems that it has to be constructed "by hand", and then, by counting dimensions, it happens to be an isomorphism... Anyways, thanks for clearing that up, I'll edit my question. Regards, –  Bruno Joyal May 24 '13 at 22:01
2  
Eh. I think the claim that there is a canonical isomorphism between $\Lambda(V^{\ast})$ and $(\Lambda V)^{\ast}$ is a little strong. I can think of two such isomorphisms one might try to write down, and I have some suspicion I know which one is the "right" one, but I'm not entirely convinced. –  Qiaochu Yuan May 24 '13 at 22:42
2  
@QiaochuYuan, I am using the word canonical in the precise sense of natural. There can be several natural isomorphisms between functors. The word canonical has traditionally an added nuance that points to a some sort of preferrence, but that ony makes sense in a context; in other words, right is context-dependent here as everywhere. –  Mariano Suárez-Alvarez May 24 '13 at 23:37

I would like to address the question posed in the title in its full generality, rather than discussing single cases.

Two parallel functors are simply objects in a functor category. Citing Awodey:

One way to prove that two objects are not isomorphic is to use “invariants”: attributes that are preserved by isomorphisms. If two objects differ by an invariant they cannot be isomorphic. Generalized elements provide an easy way to define invariants....

So if you are convinced that 2 functors are not isomorphic, try to use that feeling to provide invariants or generalized elements in the functor category that back up your intuition.

If the functor category is locally small, the above procedure amounts to using the Yoneda embedding: 2 objects in a locally small category are isomorphic if and only if their Yoneda embeddings are isomorphic.

This is certainly a slick and systematic procedure, but it may not be that easy.

share|improve this answer
    
This doesn't address the question. –  Martin Brandenburg May 25 '13 at 9:58
    
what is the question in your opinion @MartinBrandenburg ? –  magma May 25 '13 at 10:19
1  
Apart from the observation that one can use invariants (which was in a comment which is now deleted, I dunno why), and which applies to anything, not only functors, the rest of what you wrote is basically saying that to show that two functors are not isomorphic you need to show they are not :-) –  Mariano Suárez-Alvarez May 25 '13 at 14:34
    
@MarianoSuárez-Alvarez I was not aware of a comment mentioning invariants. I looked at this question only a few hours ago and I did not see that comment. Please note that I plainly said that this procedure is applicable to anything, as long as you can see the * any things* as objects in a suitable category. The rest of my answer is not a tautology. It is a consequence of the Yoneda lemma. I simply emphasized that: to see whether 2 objects are isomorphic, you can check whether the respective Yoneda embeddings are isomorphic. Cont.... –  magma May 25 '13 at 15:17
    
@MarianoSuárez-Alvarez..... This may or may not be easier, but is certainly a different problem, thus not a tautology. Awodey gives an explicit example of this with posets.I realize that my answer is very general, but so appears to be (some of) the OP's questions (title and 3rd paragraph). You and others have covered the specific examples mentioned in the rest of the Op's post. By the way, I do appreciate your smiley, we are all here to help and learn from each other in a relaxed atmosphere :-) –  magma May 25 '13 at 15:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.