I have seen two different definitions of the tensor product and I am asking about their relation.

  1. Definition (for modules in most abstract algebra books). Given two vector spaces $U$ and $V$ over $F$, the tensor product $T = U \otimes V$ is a vector space $T$ over $F$ together with a bilinear map $\pi : U \times V \to T$ such that every bilinear map on $U$ and $V$ factors through $T$ (by a unique linear map). This could easily be extended to the tensor product of an arbitrary but finite number of vector spaces.

  2. Definition (from Spivak, Calculus on Mainfolds and Munkres, Analysis on Mainfolds). Given a vector space $V$ over $F$, the $k$-fold tensor is the set of all multilinear maps from $V^k$ to $F$ (hence the tensor product of two spaces is the set of bilinear maps).

    What is the relation of Definition 1 and Definition 2?

The first definition gives an isomorphism $\operatorname{Bilin}(U, V, F) \cong L(U \otimes V, F)$ as the correspondence giving the unique linear map is linear. In general this gives an isomorphism between the $k$-fold multilinear maps and the linear maps from the $k$-fold tensor product to $F$ (note that we have not used the property to its full extend, as it also holds for bilinear maps not having their image in $F$, but lets restrict to that case).

But still this does not identifies the tensor product itself with the multilinear maps, but just those maps with the dual of the tensor product (which might in general not equal the tensor product itself)?

When we start from Definition 2, I am not quite sure how to show the universal property from Definition one. So lets consider the $2$-fold tensor product of $V$ over $F$, then we have to show that $\operatorname{Bilin}(U,V, F)$ fulfills this (universal) property. The only possible way that comes to my mind is to define $$ \pi(u,v) = u^{\ast} \cdot v^{\ast} $$ where $u^{\ast}$ and $v^{\ast}$ denote the correspoding dual elements of $V^{\ast}$ (after choosing a basis in $V$ and thereby having a dual basis in $V^{\ast}$ according to the usual construction, see here). And if $\varphi : U \times V \to F$ is another bilinear map, we have to show that $$ \varphi(u,v) = h(u^*\cdot v^*) $$ for a unique linear map, if we define $h$ that way (using that $u \mapsto u^{\ast}$ in injective) we could show that it is linear (using $\{ e_i^{\ast}\}$ and some computations). But in general $\{ u^{\ast} \mid u \in V \}\ne V^{\ast}$, so it is not clear to me how to extend $h$ to all of $\mbox{Bilin}(U,V,F)$, hence this only works in the case of reflexivity (which include the finite-dimensional vector spaces).

So what I have done just works for reflexive vector spaces, but am I on the right track? Maybe there are simpler arguments, and Munkres does not restrict his definition to finite-dimensional vector spaces only, so does it makes sense to define the tensor product like in definition two, but then how does it relate to definition one?


2 Answers 2


Definition 2 is actually the dual of the tensor product of $n$ copies of $V$. One can show that if $V$ is finite dimensional, then one has a canonical isomorphism $$(V\otimes\cdots\otimes V)^*\cong V^*\otimes\cdots\otimes V^*.$$ If moreover we fix a basis of $V$ (or more generally an inner product), then we have a canonical isomorphism $V\cong V^*$, so that the two definitions coincide. However, the "correct" definition is the first one.


You misunderstood what Spivak and Munkres write in their books. They define a $k$-tensor on $V$ as a multilinear map $f : V^k \to \mathbb R$. The set of all $k$-tensors turns out to be a vector space over $\mathbb R$. Spivak denotes it by $\mathfrak I^k(V)$, Munkres by $\mathcal L_k(V)$. They do not claim that this vector space is the tensor product of $k$ copies of $V$ in the sense of Definition 1 based on a universal property. Actually they do not consider this universal property at all. In other words, they do not consider the concept of a tensor product of vector spaces.

Certainly you know the standard construction of $U \otimes V$ as a quotient of the vector space $F(U,V)$ having as a basis the set $U \times V$. The bilinear map $\pi : U \times V \to U \otimes V$ such that $(U \otimes V,\pi)$ has the universal property of the tensor product is given by mapping each $(u,v) \in U \times V$ to $(u,v) \in F(U,V)$ and then projecting to $u \otimes v = [(u,v)] \in U \otimes V$.

If $\dim U =m, \dim V = n$, then $\dim U \otimes V = mn$. Thus you can take any vector space $X$ with dimension $mn$ and any linear isomorphism $h : U \otimes V \to X$ and the pair $(X, \pi' = h \circ \pi)$ will have the universal property of the tensor product. This applies in particular to $X = \mathcal L(U \times V , \mathbb R)$ = vector space of all bilinear maps $U \times V \to \mathbb R$. For $U = V$ this is nothing else than $\mathcal L_2(V)$. Unfortunately there is a serious problem: The map $\pi' : U \times V \to \mathcal L(U \times V , \mathbb R)$ does not have a "natural" description as $\pi : U \times V \to U \otimes V$, but involves the choice of a linear isomorphism $h : U \otimes V \to \mathcal L(U \times V , \mathbb R)$. This choice is arbitrary, there is no "canonical" $h$. Thus it is not a good approach to introduce the tensor product of $U, V$ as $\mathcal L(U \times V , \mathbb R)$ - we lack a canonical bilinear $U \times V \to \mathcal L(U \times V , \mathbb R)$.

However, if $U, V$ are finite-dimensional, we can take $X = \mathcal L(U^* \times V^*, \mathbb R)$ and $\pi' : U \times V \to X, \pi'(u,v)(f,g) = f(u) \cdot g(v)$. This map does not involve any choices - it is a canonical bilinear map. You can easily verify that it has the universal property of the tensor product. In other words, we can naturally identify $V \otimes V$ with $\mathcal L_2(V^*)$ in the sense of Spivak / Munkres.

Since $V^{**} \approx V$ naturally, we can also naturally identify $V^* \otimes V^*$ with $\mathcal L_2(V)$.

In the literature (especially in physics) you will find the notions of covariant and contravariant tensors associated to a vector space $V$. The vector space $V^* \otimes V^* = \mathcal L_2(V)$ is the space of covariant tensors, the vector space $V \otimes V = \mathcal L_2(V^*)$ is the space of contravariant tensors. For a flavor see Why is tensor from a vector space covariant, not contravariant?

Spivak and Munkres only consider covariant tensors.


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