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So I was studying tensor products from the book "An Introduction to Tensors and Group Theory for Physicists". After proving the fact that $ \{ e_i \otimes f_j\}_{i \in \mathcal{I}, \, j\in \mathcal{J}}$ (where $\{e_i\}_{i \in \mathcal{I}} $ and $\{f_j\}_{j \in \mathcal{J}}$ are bases for $V$ and $W$ respectively.) is a basis for the tensor product of $V$ and $W$ and hence proving that $\dim V\otimes W = \dim V \dim W $. The author then asks us to convice ourselves that the following two sets of isomorphisms are true and says that proving them is not that easy.
$$ (V \otimes W)^* \cong V^* \otimes W^*$$

and that

$$ (V_1 \otimes V_2) \otimes V_3 \cong V_1 \otimes(V_2 \otimes V_3)$$

Now, the thing that I want to ask is the following: I can certainly count the dimensions of each vector space in the question and prove the statements easily but somehow that doesn't feel right. It maybe because I was warned that the supposed proofs for these relations would be hard or it is because proving such isomorphisms by mere dimension counting doesnot provide any insight to the structure of the vector spaces at hand. Either way, I am looking for answers to these two questions

  1. Should I feel Ok to prove isomorphisms by mere dimension counting? Why? or Why not?
  2. Does there exist a "natural" isomorphism between these spaces similar to the usual isomoprhism that maps $V$ onto $(V^*)^*$ (I'm obviously talking about the map that assigns each $v \in V$ to $ev_v$).

Thanks in advance.

irmbil
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    For the first question, it is important for more abstract mathematical result to have an explicit canonical isomorphism each time, furthermore how will you do it if the dimension is infinite? For the second Yes, construct it? – julio_es_sui_glace Apr 01 '24 at 23:00
  • That's very nice to hear. Can you get me started on that? – irmbil Apr 01 '24 at 23:02
  • 2: No natural isomorphism. Each choice of basis gives you a different isomorphism. You might think that isomorphic indifference to the choice of basis might expand to irrelevance of the choice, but, no, there are just a lot of isomorphisms. Discussed further here: https://math.stackexchange.com/questions/763199/on-the-canonical-isomorphism-between-v-and-v – Eric Towers Apr 01 '24 at 23:04
  • You space is spanned by tensors, use this property to easily define them. – julio_es_sui_glace Apr 01 '24 at 23:08
  • @EricTowers He meant the iso $V \simeq V^{**}$ in the finite dimensional case – julio_es_sui_glace Apr 01 '24 at 23:10
  • The hard part is not to construct them, but to prove they are iso. – julio_es_sui_glace Apr 01 '24 at 23:21
  • @julio_es_sui_glace I think you want me to define a mapping from $(V_1 \otimes V_2) \otimes V_3$ to $V_1 \otimes (V_2 \otimes V_3)$ that just shifts the paranthesis to the right. (Basically a mapping that sends basis vectors to the basis vectors along with rest of the linear combination.) I believe this was the same exact construct that we used to prove isomorphic iff have the same dimension. – irmbil Apr 01 '24 at 23:26
  • You don't really need basis at least for the injective part, what do you send $(\sum v_1^i \otimes v_2^i)\otimes v_3$ to? – julio_es_sui_glace Apr 01 '24 at 23:27
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    I sent $\sum_ {ijk} c_{ijk} (v_1^i \otimes v_2^j) \otimes v_3^k $ to $ \sum_ {ijk} c_{ijk} v_1^i \otimes (v_2^j \otimes v_3^k) $ and it seems like this works – irmbil Apr 01 '24 at 23:53
  • For the first one, I suggest you try to find an iso from $V^* \otimes W^$ to $(V \otimes W)^$ – julio_es_sui_glace Apr 02 '24 at 10:54

1 Answers1

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  1. In each of these cases, if you are counting dimensions by choosing/constructing a basis for either side, you can construct an isomorphism by mapping the bases to each other.
  2. In each of the cases you mention, there is an "obvious" way of doing this so that the map does not actually depend on the basis, which you have also found in the "double-dual" and "associativity" examples. These will be natural in the formal sense of the word natural.

If you have a proof that $\{e_i \otimes f_j\}$ is a basis of $V \otimes W$ then I disagree with the book about proving the two isomorphisms being not easy.

ronno
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