In multilinear algebra books tensors are usually defined through the universal property. Given a family of $k$ vector spaces $V_1,\dots,V_k$ over the same field $F$ we want to construct a space $S$ and a map $T:V_1\times\cdots\times V_k\to S$ such that for every vector space $W$ and multilinear $f:V_1\times\cdots\times V_k\to W$ we have a unique linear $g:S\to W$ such that $f = g\circ T$.

We then put $S=F(V_1\times\cdots\times V_k)/S_0$ where $S_0$ is generated by all elements of $F(V_1\times\cdots\times V_k)$ with the form

$$(v_1,\dots,v_i'+v_i'',\dots,v_k)-(v_1,\dots,v_i',\dots,v_k)-(v_1,\dots,v_i'',\dots,v_k)$$ $$(v_1,\dots,kv_i,\dots,v_k)-k(v_1,\dots,v_i,\dots,v_k).$$

Then we write $S=V_1\otimes\cdots \otimes V_k$ and denote $T(v_1,\dots,v_k)=v_1\otimes \cdots \otimes v_k$. This is a good definition, motivated by an algebraic problem.

In differential geometry books, however, tensors on a vector space $V$ are defined as multilinear functions from $V^k$ to $\Bbb R$. The motivation obviously is that we are building objects that in some sense given $k$ directions is capable of giving a number being linear in each direction. They usually denote $T_k(V)$ the space of such tensors.

Now, the space $T_k(V)$ is isomorphic to $V^{\otimes k}$, so my doubt is not that. My doubt is: is there some benefit of working in differential geometry with the spaces $T_k(V)$ rather than $V^{\otimes k}$? From the geometrical point of view, is there some difference between working with $T_k(V)$ and $V^{\otimes k}$?

There's one similar question here but I didn't find an answer there.

Thanks very much in advance!


2 Answers 2


The space $T^k(V)$ is not isomorphic to $V^{\otimes k}$ but to its dual $(V^{\otimes k})^*$.
The distinction is very important if you pass from vector spaces to vector bundles on a manifold: the tangent bundle for example is not isomorphic to the cotangent bundle and this is only the case $k=1$ !
Also, I think that many authors introduce tensors through multilinear maps because they fear that genuine tensor spaces are more complicated than spaces consisting of multilinear maps, so they take advantage of the canonical isomorphism $V=V^{**}$ to replace tensors by multilinear maps.

  • $\begingroup$ "The space $T^k(V)$ is not isomorphic to $V^{\otimes k}$ but to its dual $(V^{\otimes k})^*$" - Are you sure? If $T^k(V)$ is isomorphic to $(V^{\otimes k})^*$, it's also isomorphic to $V^{\otimes k}$, since $V^{\otimes k}$ and $(V^{\otimes k})^*$ are isomorphic...isn't it? $\endgroup$
    – Filippo
    Jan 29, 2021 at 17:20

One important reason to start differential geometry with the viewpoint that tensors are multilinear maps is that the first tensors you consider arise naturally as multilinear maps. For example, the metric tensor tells you how to take the inner product of vectors in a tangent space. It might seem like an unnecessary complication to define tensors abstractly when the tensors you want to study already look like $T_{k}(V)$.


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