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We can identify spaces $V$ and $V^{**}$ by canonical isomorphism: $$A:V\to V^{**},$$ $$Av(f)=f(v),$$ for any $f\in V^*$.

But why we cannot identify $V$ and $V^*$ by $e^{*}_{i}(e_j)=\delta_{ij}$ (I understand that after change the basis of $V$ operator $B: V\to V^*$ will be changed)? What means that spaces $V$ and $V^{**}$ are identical? How we can use it?

Rudy the Reindeer
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Aspirin
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    $V$ and $V^{}$ are not "identical", but in the finite dimensional case they are canonically isomorphic. The isomorphism $A$ does not depend on the choice of basis for $V$, so it is "coordinate-free". Whereas the isomorphism between $V$ and $V^*$ is non-canonical*, because* it depends on the choice of basis. So $V$ and $V^$ are also* isomorphic, but non-canonically; there is no "natural" or "coordinate-free" isomorphism. – Arturo Magidin Jul 13 '11 at 19:46
  • I don't understand how $e_i^{\ast} (e_j) = \delta_{ij}$ defines an isomorphism from $V$ to $V^{\ast}$. – Qiaochu Yuan Jul 13 '11 at 19:49
  • The keyword is canonical. (BTW, you should add "finite dimensional" to your statement.) All $d$ dimensional vector spaces over the same base field $K$ are isomorphic. And the $e_j\mapsto e^*_j$ operation once you fix a basis give an explicit isomorphism. – Willie Wong Jul 13 '11 at 19:49
  • Isn't your isomorphism $A$ going from $V^**$ to $V$? – Fabian Jul 13 '11 at 19:50
  • @Qiaochu: Pick a basis $e_1,\ldots,e_n$, maps $e_i\mapsto e_i^$, where $e_i^$ is defined via $e_i^*(e_j)=\delta_{ij}$... – Arturo Magidin Jul 13 '11 at 19:50
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    @Fabian: No: $v\in V$, and $Av\in V^{**}$ so when you apply it to $f\in V^$, you should get an element of $K$, which you do, namely $f(v)$. He is giving the value of $Av$ at $f\in V^$, so $Av\in V^{**}$. – Arturo Magidin Jul 13 '11 at 19:52
  • @Arturo: thanks a lot. I don't know why I was confused. – Fabian Jul 13 '11 at 19:58
  • @Arturo: what means canonically isomorphic? – Aspirin Jul 13 '11 at 20:38
  • And why for canonically isomorphic spaces we use the notation $V = W$, but for non-canonical $V \cong W$ – Aspirin Jul 13 '11 at 20:46
  • @Alyushin: I wouldn't use $=$ for any isomorphism, canonical or not; I don't know why whoever introduced you to the isomorphism. The reason we say it is "canonical" is that it does not depend on how the vector space is presented (on its basis or description), but only on the fact that it is a vector space (the map does not depend on anything other than the vector space properties; the fact that it is an isomorphism only on the finite dimensionality). Regardless of how you think about the vector space, the isomorphism is the same. This is not true of the isomorphism $V\cong V^*$. – Arturo Magidin Jul 14 '11 at 06:25
  • You may also find of interest Why are vector spaces not isomorphic to their duals? – Bill Dubuque Jun 02 '12 at 15:30

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Prof Magidin has answered your question, but I wanted to mention the following exercise: suppose that you give me, for each finite-dimensional vector space $V$ (over $\mathbf{R}$, say), an isomorphism $h_V\colon V \to V^*$. Then you can show that there is some diagram $$ \begin{array}{ccc} V & \xrightarrow{f} & W \\ \downarrow & & \downarrow \\ V^* & \xleftarrow{f^*} & W^* \end{array} $$ (where the vertical arrows are $h_V$ and $h_W$) which does not commute, where $(f^*\lambda)(v) = \lambda(f(v))$ is the usual dual of a linear map. This is not to say that dualizing isn't a functor!

Dylan Moreland
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    See also the discussion in the very similar question http://math.stackexchange.com/questions/26692/can-we-say-that-v-cong-v-is-not-natural/ – Dan Petersen Dec 17 '11 at 10:51
  • Also of interest Why are vector spaces not isomorphic to their duals? – Bill Dubuque Jun 02 '12 at 15:30
  • As an addendum to what Dylan wrote, note that the evaluation morphism $\mathrm{ev}V\colon V\to (V^\vee)^\vee$ is functorial in $V$ or in other words $\mathrm{ev}{-}\colon id\to ((-)^\vee)^\vee$ is a natural transformation. Therefore for any morphism $V\to W$ the analogous diagram to the one Dylan drew, always commutes! This natural transformation exists irrespective of the dimension of $V$ and happens to be an isomorphism in the finite dimensional case. – shubhankar Jun 07 '23 at 05:23
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All the other answers are great, but I feel like this simple thing was left unsaid:

The word identical is unwise to use, because it is unclear whether it means isomorphic or naturally isomorphic.

V and V* and V** are all isomorphic of course (if they are finite dimensional), after all they have the same dimension.

What you want to say is this: V and V** are naturally isomorphic but V and V* are not! The notion of natural isomorphism is defined nicely by Dylan. If you look it up you'll find that what is actually naturally isomorphic is not V with V** but rather the ** itself. That is, the functor of double-dual on the category of vector spaces over the reals is naturally isomorphic to the identity functor.

daOnlyBG
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ebrahim
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Maybe one should note that this canonical morphism $V \to V^{**}$ is in general not an isomorphism if $V$ is not finite dimensional. Think about the space $V = \oplus_{n\geq 1} K$ for some field $K$. Here already $V^*$ is not isomorphic to $V$. In Banach space theory for example a space is called reflexive precisely if this map is an isomorphism.

I think part of your confusion seems to be that people often say "identical" when they mean canonically isomorphic. If you keep in mind that most of the time things are not "the same" but canonically (or naturally) isomorphic then this will help you not to get confused. In particular the spaces $V$ and $V^{**}$ are not identical.

mland
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