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Are these two definitions of a tensor product equivalent?

A tensor product of two vector spaces $V$ and $W$ over field $K$ is a pair $(T,\otimes)$ where $T$ a vector space over $K$ and $\otimes$ a bilinear map with the property that ...

Def 1.(ref) ... for every bilinear map $B_{L}\colon V\times W\to X$ where $X$ any vector space over $K$, there exists a unique linear map $F_{\otimes}\colon T\to X$ so that $B_{L}=F_{\otimes}\circ\otimes$.

Def 2.(ref) ... if $\{\bar{e}_{i}\}$ is a basis of $V$ and if $\{\bar{f}_{j}\}$ is a basis of $W$ then $\{\bar{e}_{i}\otimes\bar{f}_{j}\}$ is a basis of $T$ (used notation $\bar{e}_{i}\otimes\bar{f}_{j}\equiv \otimes(\bar{e}_{i},\bar{f}_{j})$)

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Yeah man, they're the same. Namely, Def. 1 is the like "moral definition", i.e. what tensor products do, and Def. 2 is how to make it. Like Def. 1: "Cookies-they taste good" Def. 2: "get some flour n sugar n stuff". –  uncookedfalcon May 24 '12 at 9:40
    
Formally, $(T, \otimes)$ is unique up to unique isomorphism, cuz it's the "initial object in vector spaces with a bilinear map from $V \times W$", so you just need to show Def.2 satisfies Def.1 with the map $\otimes(v,w) = v \otimes w$ –  uncookedfalcon May 24 '12 at 9:42
    
You need to show it both ways, no? Def.1 $\Leftrightarrow$ Def.2 –  Wox May 24 '12 at 9:59
    
That is certainly reasonable, but in this case, we only have $Def.2 \Rightarrow Def.1$. Namely, anything isomorphic to the thing in Def. 2 will also satisfy definition 1! So it's more like, Def.2 is the standard construction of Def. 1. –  uncookedfalcon May 24 '12 at 10:18
    
@Wox Here, just in case you don't know how to vote and accept answers: meta.math.stackexchange.com/a/3287/5798 –  Matt N. May 24 '12 at 12:29
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2 Answers

up vote 3 down vote accepted

Yes, they are equivalent. To show this first suppose that $(T, \otimes)$ is a tensor product of $V$ and $W$ with respect to definition 1, let $(e_i)_{i\in I}$ and $(f_j)_{j\in J}$ be bases of $V$ and $W$, respectively. Define a bilinear map $B\colon V \times W \to K^{(I \times J)}$ by $B(e_i, f_j) = \delta_{(i,j)}$, where $\delta_{(i,j)}$ denotes the element of $K^{(I\times J)}$ which maps $(i,j)$ to 1 and all other pairs to 0. Then there is a unique linear $F\colon T \to K^{(I\times J)}$ with $F \circ \otimes = B$. We have $F(e_i \otimes f_j) = \delta_{(i,j)}$ for all $(i,j) \in I \times J$, so $F$ is a surjection. Moreover, as the $\delta_{(i,j)}$ are linearly independent, so are the $e_i \otimes f_j$. Let $U$ be the subspace of $T$ spanned by $(e_i\otimes f_j)_{i,j}$, let $X$ be a complement of $U$, i. e. $T = U \oplus X$. Let $\phi\colon X \to K^{(I \times J)}$ be a linear map, and define $G\colon T \to K^{(I \times J)}$ by $G(u+x) = F(u) + \phi(x)$ for $u \in U$ and $x \in X$. Then for $v \in V$, $w \in W$ as we have $v \otimes w \in U$ it holds $G(v \otimes w) = F(v \otimes w) = B(v,w)$. By uniqueness of $F$ we must have $G = F$ for all $\phi$. This yields $X = 0$, so $U = T$ and as $(e_i \otimes f_j)$ is a basis of $U$, it is one of $T$.

Now suppose $(T,\otimes)$ fulfills definition 2. Let $B\colon V \times W \to X$ be bilinear. Let $(e_i)$ resp. $(f_j)$ be bases of $V$ resp. $W$. Then $(e_i \otimes f_j)$ is a basis of $T$, so $F(e_i \otimes f_j) := B(e_i, f_j)$ extends uniquely to a linear mapping $F\colon T \to X$, which obviously has $F \circ \otimes = B$. If $G\colon T \to X$ is linear with $G \circ \otimes = B$ then we have $G(e_i \otimes f_j) = B(e_i, f_j) = F(e_i \otimes f_j)$, but now, as two linear mappings agreeing on a basis are identical we have $F= G$, so $F$ is unique.

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Thanks for the elaborate answer! –  Wox May 24 '12 at 13:54
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Recall that for linear maps $f : V \to W$, if we have a basis $\{ v_i \}$ for V and know all of the values $w_i = f(v_i)$, then that is enough to determine the value of $f(v)$ for every vector $v \in V$. Conversely, any choice we make for the $w_i$ gives us a linear map.

The main thing to note here is that we have a similar fact for bilinear maps! If $f : U \times V \to W$ is bilinear, we have bases $\{ u_i \}$ and $\{ v_j \}$, and we know the values $w_{ij} = f(u_i, v_j)$, that is enough to determine every value $f(u,v)$. Conversely, any choice for the $w_{ij}$'s gives us a bilinear map.

But now notice the similarity of the previous two paragraphs. If we had a new based* vector space $T$ whose basis vectors were labelled with pairs $(i,j)$, then picking a bilinear map from $U \times V \to W$ is the same thing as picking a linear map $T \to W$!

Once you recognize this, all of the rest is just grinding out the details.

*: a based vector space is just a vector space for which we've chosen a particular basis.

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Nicely put and very helpful to me. Thanks! –  Wox May 24 '12 at 14:43
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