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Let $V_1, ..., V_k$ be finite-dimensional vector spaces. A tensor product is defined to be the quotient space $U/I$, where $U$ denotes the free vector space generated by elements of $V_1 \times ... \times V_k$, and $I$ the subspace of $U$ generated by elements of the form

$a(v_1, ..., v_k) - (v_1, ..., av_i, ..., v_k)$

$(v_1, ..., v_i + w_i, ..., v_k) - (v_1, ..., v_i, ..., v_k)- (v_1, ..., w_i, ..., v_k)$

and the space $U/I$ is denoted $V_1 \otimes...\otimes V_k$.

Question:

How to establish an isomorphism between $(V_1 \otimes ... \otimes V_l)\otimes(V_{l+1} \otimes ... \otimes V_k)$ and $V_1 \otimes...\otimes V_k$?

This question might look silly, but I find it troublesome when dealing with the definitions above...

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    $\begingroup$ Can you do the case $k = 3$ and $l = 1$? You don't write anything about the universal mapping property of tensor products. Are you comfortable proving any other isomorphism theorems about tensor products (e.g., $V \otimes (W \oplus W') \cong (V \otimes W) \oplus (V \otimes W')$)? $\endgroup$ – KCd May 26 '13 at 13:13
  • $\begingroup$ @KCd Thanks for reminding me of the universal mapping property. But is there a straightforward way to deal with the definitions? I tried to define mappings from $V_1 \otimes V_2 \otimes V_3$ to $(V_1 \otimes V_2) \otimes V_3$ on the basis, then extend it linearly. Is it possible to see the isomorphism in this way? $\endgroup$ – Ezra May 26 '13 at 13:23
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    $\begingroup$ The universal property is the straightforward way! You can work with the generators if you like, but bear in mind that you have to respect the relations as well. $\endgroup$ – Zhen Lin May 26 '13 at 13:34
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    $\begingroup$ Zhen is absolutely correct. You do not want to prove isomorphisms between tensor product spaces and other vector spaces (including other tensor product spaces) using bases, unless you're a physicist I guess. :) You shouldn't avoid trying to use the univ. mapping property but figure out how to use it, because that's the only way you're going to get so comfortable with it that you become unafraid of that (mathematically) essential feature of tensor products. $\endgroup$ – KCd May 26 '13 at 14:14
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The definition you mention is one possible realization of the tensor product, but the “universal property” is what makes it possible to manage tensor products.

The tensor product of the vector spaces $V_1$, $V_2$, $\dots$, $V_k$ can be defined as a pair $(V,\alpha)$, where $V$ is a vector space and

$$\alpha\colon V_1\times V_2\times\dots\times V_k\to V$$

is a multilinear map such that, for any multilinear map

$$\beta\colon V_1\times V_2\times\dots\times V_k\to W$$

there exists a unique linear map $f\colon V\to W$ such that $f\circ\alpha=\beta$.

It's quite easy to show that your $U/I$, together with the map

$$\gamma\colon V_1\times V_2\times\dots\times V_k\to U/I$$

defined by

$$\gamma(v_1,v_2,\dots,v_k)=v_1+v_2+\dots+v_k+I$$

has the above property (where $v_1+v_2+\dots+v_k$ denotes the element $(v_1, v_2, \ldots, v_k)$ of $U$).

If $(V,\alpha)$ is another “realization" of the tensor product, then, by the universal property, there are $f\colon V\to U/I$ and $g\colon U/I\to V$ such that

$$f\circ\alpha=\gamma,\qquad g\circ\gamma=\alpha$$

and it's easy to prove that $f\circ g$ and $g\circ f$ are both identity maps. Namely,

$$(g\circ f)\circ\alpha=g\circ(f\circ\alpha)=g\circ\gamma= \alpha=1_V\circ\alpha$$

so, by the uniqueness, $g\circ f=1_V$. Similarly for $f\circ g$.

In this way you can denote any realization of the tensor product by $V_1\otimes V_2\otimes\dots\otimes V_k$. The multilinear map is not expressed, but it is implicit in the choice of the realization; conventionally the map is denoted by $(v_1,v_2,\dots,v_k)\mapsto v_1\otimes v_2\otimes\dots\otimes v_k$.

Any element of $V_1\otimes\dots\otimes V_k$ is a sum of elements of the form $v_1\otimes v_2\otimes\dots\otimes v_k$. For this you can use abstract reasoning or the explicit realization.

Now your problem boils down to showing that $$ V_1\otimes V_2\otimes\dots\otimes V_k\otimes V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l $$ is a realization of the tensor product $$ (V_1\otimes V_2\otimes\dots\otimes V_k)\otimes (V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l) $$ via a suitable bilinear map. The bilinear map $$ (V_1\otimes V_2\otimes\dots\otimes V_k)\times (V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l)\to V_1\otimes V_2\otimes\dots\otimes V_k\otimes V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l $$ is just $$(v_1\otimes\dots\otimes v_k,v_{k+1}\otimes\dots \otimes v_l)\mapsto v_1\otimes\dots\otimes v_k\otimes v_{k+1}\otimes\dots\otimes v_l $$ extended by bilinearity.

You can have some doubts about the last map being well defined. But remember you have a given multilinear map $$ \alpha\colon V_1\times\dots\times V_k\times V_{k+1}\times\dots\times V_l \to V_1\otimes\dots\otimes V_k\otimes V_{k+1}\otimes\dots\otimes V_l = T $$

Define, for any $y\in V_{k+1}\times\dots\times V_l$, the map $\beta_y\colon V_1\times\dots\times V_k\to T$ by $\beta_y(x)=\alpha(x,y)$. This map is multilinear, so it induces a unique linear map $$f_y\colon V_1\otimes\dots\otimes V_k\to T$$ such that $f_y\circ\alpha_k=\beta_y$, where $\alpha_k : V_1 \times \cdots \times V_k \to V_1 \otimes \cdots \otimes V_k$ is again the canonical multilinear map.

Thus you have a multilinear map $y\mapsto f_y$ $$ V_{k+1}\times\dots\times V_l\to \mathrm{Hom}(V_1\otimes\dots\otimes V_k,T) $$ that induces a unique linear map $$ g\colon V_{k+1}\otimes\dots\otimes V_l\to \mathrm{Hom}(V_1\otimes\dots\otimes V_k,T) $$ The bilinear map you're looking for is just $$h(w_1,w_2)=g(w_2)(w_1)\qquad (w_1\in V_1\otimes\dots\otimes V_k, w_2\in V_{k+1}\otimes\dots\otimes V_l).$$

Indeed, if $w_1=v_1\otimes\dots\otimes v_k$ and $w_2=v_{k+1}\otimes\dots\otimes v_l$, showing that $$ h(w_1,w_2)=\alpha(v_1,\dots,v_k,v_{k+1},\dots,v_l) $$ is just a tedious verification.


Final comments

Proving the existence of the isomorphism with the “explicit” realization you start with is time consuming and not worth the trouble. The explicit realization $U/I$ is only used to show that a tensor product exists. Then you can forget about it and use only the universal property.

This is similar to how one does when constructing the real numbers from the rationals. The existence of a field with the required properties can be proved with Dedekind cuts or Cauchy sequences or whatever; then one proves that a field with those properties is unique up to isomorphisms and the actual construction is not used any more: one just employs the properties of the field.

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  • $\begingroup$ Thanks for your detailed answer! I still have a question: how can I see that the map $(v_1\otimes...\otimes v_k, v_{k+1} \otimes ... \otimes v_l) \mapsto v_1\otimes...\otimes v_k \otimes v_{k+1} \otimes ... \otimes v_l$ is well-defined? $\endgroup$ – Ezra May 26 '13 at 13:50
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    $\begingroup$ That is a good question. Certainly that formula on its own is not really a complete definition of a bilinear mapping (for someone first learning this stuff). Here is an idea. Let $V, V', V'', W$ be vector spaces (over the same field) and suppose we have a trilinear function $f \colon V \times V' \times V'' \rightarrow W$. For $v \in V$, let $f_v \colon V' \times V'' \rightarrow W$ by $f_v(v',v'') = f(v,v',v'')$. Then each $f_v$ is bilinear, and use it to get a linear map $\varphi_v \colon V' \otimes V'' \rightarrow W$ where $\varphi_v(v' \otimes v'') = f(v,v',v'')$. That is [to be contd.] $\endgroup$ – KCd May 26 '13 at 14:22
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    $\begingroup$ the "mapping" $V \times (V' \otimes V'') \rightarrow W$ where $(v,v' \otimes v'') \mapsto f(v,v',v'')$ is well-defined. Check it's bilinear! We have condensed a trilinear map into a bilinear map. You can take it a step further and get a linear map $V \otimes (V' \otimes V'') \rightarrow W$ with effect $v \otimes (v' \otimes v'') \mapsto f(v,v',v'')$ on elementary tensors. If your original map had been $V \times V' \times V'' \rightarrow (V \otimes V') \otimes V''$ by $f(v,v',v'') = (v \otimes v') \otimes v''$ then you'd now have a linear map between realizations of a triple tensor product. $\endgroup$ – KCd May 26 '13 at 14:28
  • $\begingroup$ @Ezra I've added the proof. $\endgroup$ – egreg May 26 '13 at 14:30

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