# “A gentleman never chooses a basis.”

Around these parts, the aphorism "A gentleman never chooses a basis," has become popular.

Is there a gentlemanly way to prove that the natural map from V to V** is surjective if V is finite dimensionsal?

As in life, the exact standards for gentlemanliness are a bit vague. Some arguments seem to be implicitly pick basis. I'm hoping there's an argument which is unambigously gentlemanly.

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I'm having trouble coming up with a sufficiently patriarchal argument. Does "these parts" refer to the pre-suffrage era? –  S. Carnahan Oct 13 '09 at 3:31
That's fair. I should have gone for something more gender neutral. Although "gentlemanly/ladylike" is a bit awkward, and something like "classy" doesn't have the same anachronistic feel. Any suggestions? –  Richard Dore Oct 13 '09 at 18:50
My personal preference is to avoid any reference to gender or class (or indeed membership in any group associated to historical persecution - e.g., we don't say that bases are for Jewish or homosexual people). This may make your question seem less colorful, but I think it is worthwhile to make mathematics more welcoming to people of all kinds. If you're still looking for an obnoxious elitist tone, I suggest replacing "gentleman" with "true mathematician" and "gentlemanly" with "mathematically cultured". –  S. Carnahan Oct 14 '09 at 1:55
In my mind, "gentleman" refers to politeness rather than social class, but I can see where the problem comes from. Perhaps a good alternative is "my mommy said it's not polite to choose a basis." My mom didn't tell me that, so as a kid, I chose bases left and right; now I regret it. –  Anton Geraschenko Oct 14 '09 at 14:33
This doesn't constitute a proof, but: Suppose that the result of a certain proof looks obvious in notation A, but deep and mysterious in notation B. This is usually a reason to prefer notation A. In Penrose's abstract index notation, which doesn't require a choice of basis, mapping one-dimensional space V to V* takes element $x_a$ to element $x^a$. If you then continue with V* to V**, you take $x^a$ to (drumroll, plese) $x_a$. If the mapping from V to V** wasn't surjective (and, in fact, an isomorphism) then abstract index notation would be inconsistent. –  Ben Crowell Sep 18 '12 at 4:12

Following up on Qiaochu's query, one way of distinguishing a finite-dimensional $V$ from an infinite one is that there exists a space $W$ together with maps $e: W \otimes V \to k$, $f: k \to V \otimes W$ making the usual triangular equations hold. The data $(W, e, f)$ is uniquely determined up to canonical isomorphism, namely $W$ is canonically isomorphic to the dual of $V$; the $e$ is of course the evaluation pairing. (While it is hard to write down an explicit formula for $f: k \to V \otimes V^*$ without referring to a basis, it is nevertheless independent of basis: is the same map no matter which basis you pick, and thus canonical.) By swapping $V$ and $W$ using the symmetry of the tensor, there are maps $V \otimes W \to k$, $k \to W \otimes V$ which exhibit $V$ as the dual of $W$, hence $V$ is canonically isomorphic to the dual of its dual.

Just to be a tiny bit more explicit, the inverse to the double dual embedding $V \to V^{**}$ would be given by

$$V^{\ast\ast} \to V \otimes V^* \otimes V^{\ast\ast} \to V$$

where the description of the maps uses the data above.

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OK, great! So you can define finite-dimensionality without mentioning bases (or chains of subspaces). The answer to the question is then easy. But this recasting of the definition of finite-dimensionality is, I think, much the most interesting thing. –  Tom Leinster Oct 21 '09 at 22:54
Yes, there a number of ways one might think of characterizing finite-dimensionality (including being isomorphic to its double dual!), Noetherian/Artinian hypotheses, etc. But some of these characterizations don't port so well to modules over other commutative rings. The present characterization is equivalent to being finitely generated and projective, for any commutative ring. –  Todd Trimble Oct 22 '09 at 4:15
When you say "isomorphic to its double dual" you presumably mean its algebraic double dual. –  Loop Space Nov 8 '09 at 21:26
Presumably Andrew means that one almost never talks about unadorned infinite-dimensional vector spaces. An analyst naturally thinks of the dual of a finite-dimensional vector space as a special case of the continuous dual of a topological vector space, and in this situation spaces are rarely isomorphic to their double duals. –  Qiaochu Yuan Nov 23 '09 at 15:24
I guess Andrew also means that, for example, Hilbert spaces <em>are</em> isomorphic to their continuous double duals. –  Qiaochu Yuan Nov 23 '09 at 15:26

At the price of being too categorical for the question, one can follow up Todd's answer as follows. Consider any closed symmetric monoidal category $\mathcal{V}$ with product $\otimes$ and unit object $k$, such as vector spaces over a field $k$. Let $V$ be an object of $\mathcal{V}$ and let $DV = Hom(V,k)$. Just from formal properties of $\mathcal{V}$, there are canonical maps $\iota\colon k\to Hom(V,V)$ and $\nu\colon DV\otimes V\to Hom(V,V)$, which are the usual things for vector spaces. Say that $V$ is dualizable if there is a map $\eta\colon k\to V\otimes DV$ such that $\nu \circ \gamma \circ \eta = \iota$, where $\gamma$ is the commutativity isomorphism. Formal arguments show that $\nu$ is then an isomorphism and if $\epsilon\colon DV\otimes V \to k$ is the evaluation map (there formally), then, with $W=DV$, $\eta$ and $\epsilon$ satisfy the conditions Todd stated for $e$ and $f$. This is general enough that it can't have anything to do with bases. But restricting to vector spaces, we can choose a finite set of elements $f_i\in DV$ and $e_i\in V$ such that $\nu(\sum f_i\otimes e_i) = id$. Then it is formal that $\{e_i\}$ is a basis for $V$ with dual basis $\{f_i\}$. This proves that $V$ is finite dimensional, and the converse is clear as in Todd's answer. There is a result in Cartan-Eilenberg called the dual basis theorem that essentially points out that the precisely analogous characterization holds for finitely generated projective modules over a commutative ring $k$, with the same proof.

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Yes, this is a nice argument, Peter. –  Todd Trimble Sep 17 '12 at 18:52
To be pedantic, in the case of f.g. projective modules over a commutative ring, "dual basis theorem" is a slightly unfortunate name, since neither $\{e_i\}$ or $\{f_i\}$ are necessarily bases of $V$ or $DV$. –  Peter Samuelson Sep 17 '12 at 22:54

Perhaps it would be most appropriate to answer your question with another question: how do you distinguish a finite-dimensional vector space from an infinite-dimensional one without talking about bases?

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Every increasing (or decreasing) sequence of subspaces stabilizes in finitely many steps. –  Richard Dore Oct 14 '09 at 4:57
Let me suggest the following strategy, then: to any chain of subspaces in V there is associated a dual chain in V*. If one can show that strict inclusions are sent to strict inclusions, then V and V* have the same dimension. –  Qiaochu Yuan Oct 15 '09 at 18:22

Work in modules over a commutative unital ring R, why not.

There is a natural transformation from the identity functor to the double dual functor, Hom(Hom(-,R),R). This is described in the other comments.

This is easily verified to be an isomorphism when restricted to R. At this stage we use a basis consisting of the element 1, but that doesn't involve any choices.

Both functors commute with finite sums, so the natural transformation is an isomorphism for all finitely generated free modules.

Since isomorphisms are closed under retracts we also obtain that our map is an isomorphism for all finitely generated projectives.

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But then you have to prove that any vector space is a free module over a field, which sounds a lot like proving the existence of a basis. By the way, why did you make this answer community wiki? –  Anton Geraschenko Oct 16 '09 at 16:14
@Josh: you don't need to make your posts community wiki for people to be able to add comments. Community wiki means (1) you don't earn any reputation from the post, and (2) other people can edit it if they have 100 reputation (if it's not community wiki, they need 2000 reputation). –  Anton Geraschenko Oct 22 '09 at 1:10

there is a canonical map $ev:V \to V^{**}$ defined by $ev(v)(\phi) = \phi(v)$. to check that it is an isomorphism in the finite dimensional setting you can just check that it is injective and this is evident from the definition.

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How do you know it's surjective though? You have to know they're the same dimension. I don't know how to prove that without getting dirty with a basis. –  Richard Dore Oct 13 '09 at 18:46

Some kind of solution proposal:

Let V be a n-dimensional vector space over a field (or a free R-module, where R is a commutative unital ring).

There is a morphism V tensor V* to End(V), which sends each v tensor lambda to the endomorphism of V that sends each w to lambda(w)v. It is an epimorphism since it's image are all finite rank endomorphisms, so it's surjective. It is a monomorphism as you can check by calculation. So this is an isomorphism.

We can calculate the dimensions: dim(V tensor V* ) = dim(End(V)), where dim(V tensor V* ) = dimV * dimV* and dim(End(V))=(dimV)^2. So the result is n * dimV* = n^2 and we get dimV* = n = dimV.

Now notice that every short exact sequence in our category splits. That implies for every monomorphism V to W, that W is isomorphic to a direct sum of V and W/V and therefore we have a dimension formula dimV + dim(V/W) = dim W. We get the result that every monomorphism from V to W with dimW=dimV is an isomorphism.

Look at the linear map ev : V to V**, which sends v to ev_v : (lambda mapsto lambda(v)), the evaluation-at-v-map. Now we make an induction: for dimV=0, the map ev is trivially an isomorphism. For dimV=n, the kernel of ev is a subspace, so we have V = ker(ev) + W with some complement W and either ker(ev)=V or ker(ev)=0 or the two subspaces have strictly smaller dimension. That would mean, by induction hypothesis, that their evaluation map, which is the restriction of the evaluation map of V, has no kernel and so we get ker(ev)=0. The case ker(ev)=V remains, where we get that V*=0 which contradicts n=dimV=dimV*.

Now ev is a monomorphism and dim(V** )=dim(V* )=dim(V), therefore ev is an isomorphism. One can check easily that this is "functorial", that is: we have a natural transformation from the identity functor to the bidual functor.

One could object that I have chosen an arbitrary flag, when I take the complement of the kernel in the induction step... but I guess without that you wouldn't use the "free" property of the modules in question, and for non-free modules there are counter-examples.

If I did something wrong, please tell me.

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How do you prove that dim(End(V)) = n^2 without choosing a basis? –  Qiaochu Yuan Oct 23 '09 at 5:31

Over real or complex (or other similar) field, where we know that for a finite-dimensional vector space all reasonable vector-space topologies coincide... V is dense in V** in the weak topology, hence in all topologies, but the (unique) topology is also complete, so V = V** (I think this works and avoids choosing a basis. Of course you would have to prove those other facts also without choosing a basis.)

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I'm sorry if this should be a comment rather than answer. It is an addendum to my previous answer. I should have pointed out that, still in a general symmetric monoidal category, if $V$ is dualizable, then a formal argument also shows that the canonical map $V \to V^{**}$ (again defined formally) is an isomorphism. Also, in answer to Peter Samuelson, while the name dual basis theorem'' dates from long before my time, it does have some justification. When $\mathcal{V}$ is modules over a commutative ring $k$, if one takes a dualizable $V$ and constructs the free module $F$ on basis $\{d_i\}$ in 1-1 correspondence with the $e_i$ in my previous post, then $\alpha(v) = \sum f_i(v) d_i$ specifies a monomorphism $\alpha\colon V\to F$ such that $\pi\alpha = id$, where $\pi(d_i) = e_i$. This completes the proof that dualizable implies finitely generated projective, with a relevant basis in plain sight.

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This can be added to your previous answer, if you like. –  David Roberts Sep 18 '12 at 2:57
Fine with me. I'm not adept at adding things or changing things, as I'm sure you have noticed. Thanks. –  Peter May Sep 18 '12 at 3:39