In the book An introduction to manifolds by Tu, Loring W, it defines the $k$-tensor as follows

Denote by $V^{k}=V \times \cdots \times V$ the Cartesian product of $k$ copies of a real vector space $V $. A function $f: V^{k} \rightarrow \mathbb{R}$ is $ k $-linear if it is linear in each of its $ k $ arguments: $$ f(\ldots, a v+b w, \ldots)=a f(\ldots, v, \ldots)+b f(\ldots, w, \ldots) $$ for all $ a, b \in \mathbb{R} $ and $ v, w \in V $. Instead of $2$-linear and $3 $-linear, it is customary to say "bilinear" and "trilinear." A $k $-linear function on $V$ is also called a $ k $-tensor on $ V $. We will denote the vector space of all $ k $-tensors on $ V $ by $ L_{k}(V) $. If $f $ is a $ k $ -tensor on $ V $, we also call $ k $ the degree of $ f $.

However in Wikipedia, it defines the $(p,q)$-tensor as follows

In this approach, a type $(p,q)$ tensor $T$ is defined as a multilinear map, $$ T: \underbrace{ V^* \times\dots\times V^*}_{p \text{ copies}} \times \underbrace{ V \times\dots\times V}_{q \text{ copies}} \rightarrow \mathbf{R}, $$ where $V^*$ is the corresponding dual space of covectors, which is linear in each of its arguments.

It is obvious that the definition of a $k$-tensor in An introduction to manifolds is just the $(0,q)$-tensors in the definition of Wikipedia.

My question is, why do not we define the tensors to be $(p,q)$-tensor in Tu, Loring W?

EDIT: I have noticed that there are no $(p,q)$-tensors in the whole book of Loring Tu.

Is it because that the elements from dual space will never be used in tensors in manifolds in the following study (so $p$ always equals to $0$)? If not, where do we use $(p,q)$-tensors in the study of manifolds? (maybe it will appear in tensor fields?)

I would really appreciate it if anyone could tell me the connections of these things.

  • 1
    $\begingroup$ Identifying $V\cong V^{**}$, vectors become $(1,0)$-tensors. Also, linear operators are equivalent to $(1,1)$-tensors. In general a $k$-linear map $V^{\times k}\to V$ can be identified with a $(1,k)$-vector. My notes might be helpful. $\endgroup$
    – Ivo Terek
    Nov 24, 2020 at 3:52
  • $\begingroup$ Another example is the Riemann curvature tensor which can be thought of as a $(1,3)$-tensor field (note that people will often say tensor when they mean tensor field). You can find details in Loring Tu’s Differential Geometry, which is a sequel to the book you’re currently reading. $\endgroup$ Jan 21, 2021 at 5:17
  • $\begingroup$ Could you please write about that more concretely in an answer? $\endgroup$
    – FFjet
    Jan 21, 2021 at 6:07

1 Answer 1


Note that even Wikipedia's definition is not the most general. The most general definition (that I know of) for a (real) tensor is a multilinear map $$ T: V_1 \times V_2 \times \cdots \times V_k \rightarrow \mathbb{R},$$ where $V_1, \dots, V_k$ are (not necessarily the same) vector spaces. And such generalized tensors definitely do appear in differential geometry and topology.

I also want to note that these simple definitions are also very misleading in the sense that the tensors that arise in geometry and topology all have symmetries or anti-symmetries in the indices. This leads to most of the headaches when learning how to deal with tensors. In fact, if you get serious with tensors, the notation can become quite a challenge. Penrose developed a particularly creative approach to dealing with this.

However, when you're first learning differential geometry and topology, almost all of the tensors encountered are indeed $(0,q)$ tensors. That's because either you're working with only exterior differential forms or you're working with a Riemannian manifold, where the metric can be used to convert any $(p,q)$-tensor into $(0,p+q)$-tensor.

It's good to know the definitions of general tensors (as described above) and $(p,q)$ tensors, because they are sometimes necessary or useful. But it's probably best to avoid devoting much attention to them in introductory courses and textbooks.

ADDED: An extremely useful canonical isomorphism is $$ \operatorname{Hom}(V,W) = V^*\otimes W $$ where $V$ and $W$ are vector spaces and $\operatorname{Hom}(V,W)$ is the space of linear maps from $V$ to $W$. Used with care, this means that you can view a $(p,q)$ tensor $T \in V_1^*\otimes\cdots\otimes V_q^*\otimes V_{q+1} \otimes \cdots \otimes V_{p+q}$ as a multilinear map $$ T: V_{1}\times\cdots\times V_{q} \rightarrow V_{q+1}\otimes\cdots\otimes V_{p+q}. $$ In short, to avoid working with $(p,q)$-tensors directly, it can be useful to view them as maps from $(q,0)$-tensors to $(p,0)$-tensors instead. Or, equivalently, as $(V_{q+1}\otimes\cdots\otimes V_{p+q})$-valued $(0,q)$-tensors.

  • 1
    $\begingroup$ Couldn't have said it better myself. +1 $\endgroup$
    – K.defaoite
    Jan 21, 2021 at 5:14
  • $\begingroup$ @K.defaoite: Couldn't have commented better myself. +1 $\endgroup$
    – timur
    Jan 21, 2021 at 5:27
  • $\begingroup$ Do we need $(p,q)$-tensors for General Relativity? $\endgroup$
    – littleO
    Jan 21, 2021 at 8:10
  • 1
    $\begingroup$ @littleO, strictly speaking, since there is a non-degenerate metric in general relativity, you can still "lower all the indices" and convert any $(p,q)$-tensor into a $(0,p+q)$-tensor, But, in any given calculation, this may or may not be a good idea. $\endgroup$
    – Deane
    Jan 21, 2021 at 15:56

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