In Physics and Differential Geometry usually tensors of type $(k,l)$ on a vector space $V$ over $\mathbb{F}$ are defined as multilinear functions

$$f : \underbrace{V\times\cdots\times V}_{k \ \mathrm{terms}}\times\underbrace{V^\ast\times\cdots\times V^\ast}_{l \ \mathrm{terms}}\to\mathbb{F}$$

this makes it quite simple to gather some understanding of $(k,0)$ tensors from one intuitive point of view. They are just $k$-linear functions of vectors and can be used like linear functions of vectors or like inner products and so on. Also it is not hard to see why one would care about these.

Now, on the other hand, tensors of type $(0,l)$ also appear in Physics quite frequently. Indeed Maxwell's Stress Tensor is:

$$\mathcal T = \epsilon_0\left[ \mathbf{E}\otimes\mathbf{E}+c^2\mathbf{B}\otimes\mathbf{B} -\frac12\sum_i\mathbf{e}_i\otimes\mathbf{e}_i\left(E^2+c^2 B^2\right) \right].$$

Those objects are not much intuitive IMHO. First, a tensor of type $(0,l)$ is a function of linear functionals in this approach, and this makes it a little bit harder to make sense from a physical and geometrical point of view.

The other possible approach to tensors is the one based on the universal property. As far as I understand, the basic idea of this approach is that in the end the construction (with quotient spaces and so on) shows that there exists a way to make sense of the product $v_1\otimes\cdots \otimes v_k$ and that it has all the nice properties we would want.

In that case, a tensor of type $(0,l)$ as defined above is an element of $V\otimes\cdots\otimes V$. Of course to understand those objects, it suffices to understand for $v,w\in V$ how $v\otimes w$ can be understood.

So my question is: I know the constructions are isomorphic and I know from a rigorous point of view what $v\otimes w$ is, now how can one intuitively make sense of $v\otimes w$? Again, thinking of it as a function of linear functionals doesn't seem much intuitive. So, just regarding it as an element of $V\otimes V$ how can we give to it some geometric and physical intuition?

The object $v\wedge w \in V\wedge V$ has one nice way to be understood: it can be thought of as the paralelogram generated by $v$ and $w$, that is one oriented area in the same way as $v$ and $w$ are oriented segments. Now, is there a nice way to understand $v\otimes w$ too?

  • $\begingroup$ In terms of matrces, it is common to identify $v \otimes w$ with either the Kronecker product or outer product. $\endgroup$ Apr 7, 2015 at 17:41

2 Answers 2


$v\otimes w$ is either of the following:

The rank 1 linear mapping $V^*\to V$ given by $\alpha\mapsto v.\langle \alpha,w\rangle$ where $\langle\;,\;\rangle:V^*\times V\to \mathbb F$ is the duality. This is the easiest to visualize.

The decomposable bilinear form $V^*\times V^*\to \mathbb F$ given by $(\alpha,\beta)\mapsto \langle\alpha,v\rangle \langle\beta,w\rangle$.

  • $\begingroup$ I think it should rather be $\alpha\mapsto w.\langle\alpha,v\rangle$, no? In which order do we supply arguments to $v\otimes w$ ? (BTW, it's so great to learn from you on MO and here, just as it was great to see you in the workshop at Ohope beach, NZ a couple of years ago :) $\endgroup$
    – rych
    Jul 1, 2015 at 5:01
  • $\begingroup$ This is up to convention. You are free to choose, but stay consistent after! $\endgroup$ Jul 1, 2015 at 17:35

First, I think it's helpful to think of a "function of a linear functional" and a "vector" as more or less equivalent, as you mentioned. A "function of a linear functional" is something that, when paired with a covector, produces a scalar. A vector fits this description exactly. Also, the dual space of a dual space is naturally isomorphic to the original space, $V^{**} \simeq V$, further hinting that the two are equivalent in some sense.

For visualization, simpler tensors already have commonplace visualizations. Scalars are just numbers. $(0,1)$-tensors can be thought of as arrows (vectors). $(1,0)$-tensors can be thought of as planes or level sets (covectors). $(1,1)$-tensors can be thought of as transformations between vectors and covectors.

You can think of higher-rank tensors as objects that will eventually become the lower-rank tensors you are familiar with, when given the right inputs. There are two main ways you can think of tensors:

  1. A machine that fills all its input slots with vectors and covectors and produces a real number.
  2. A machine that partially fills its input slots with vectors and covectors to produce a new lower-rank tensor.

For example, a tensor $T \in V \otimes V^*$ can be thought of as a machine that eats a vector $v \in V$ and a covector $\alpha \in V^*$ to produce a scalar, $T: V \otimes V^* \rightarrow \mathbb{R}$. This is viewing $T$ as a bilinear form.

We can also view $T$ as a linear transformation if we give it a single vector $v\in V$. What remains is a machine $S:V^* \rightarrow \mathbb{R}$ that takes a covector and produces a real number, i.e. a vector. So we can view $T$ as something that takes a vector to produce a new vector $T:V \rightarrow V$.

We could also interpret $T$ as a linear transformation on covectors $T: V^* \rightarrow V^*$ for the same reason.

Generally, feeding a vector to a tensor will produce a tensor with one less covariant index. Likewise, feeding a tensor a covector will produce a tensor with one less contravariant index.

This might seem backwards, but it makes sense because the space a tensor lives in and the space of vectors/covectors it accepts are "flipped". (A tensor $T\in V \otimes V \otimes V \otimes V^* \otimes V^*$ accepts inputs from the space $V^* \times V^* \times V^* \times V \times V$, since vectors and covectors must come in pairs to produce a scalar.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .