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We all know that some set of scalar functions $f$ can be in a vector space, for example a Hilbert space, or some finite-dimensional space like the vector space of all polynomials whose degree is $\leq n$. And, alongside this, is very common to hear the phrase "All vectors are tensors of rank 1".

This would made possible for one to infer: "All functions that belong to some vector space $V$ are vectors and hence are tensors". But the idea of a scalar function or a polynomial being a tensor isn't very much clear to me, since we know that vectors must hold specific rules of transformation under a coordinate change.

What am i missing?

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  • $\begingroup$ You can certainly have functions which "are tensors". For example, you can naturally regard $L^2(\mathbb{R}^2)$ as $L^2(\mathbb{R})\otimes L^2(\mathbb{R})$. $\endgroup$ – EuYu Dec 5 '17 at 0:31
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The terms "vector" and "tensor" are both relative, and the way they are commonly used does not make this clear. The thing they are relative to is a choice of "base" vector space $V$. A vector relative to $V$ is an element of $V$, and a tensor relative to $V$ of type $(m, n)$ is an element of the tensor product $V^{\otimes m} \otimes (V^{\ast})^{\otimes n}$.

When we talk about vectors or tensors transforming in some way under change of coordinates, we are talking about coordinates for $V$. So, yes, if you take $V$ to be some space of functions, it is technically true that every function in that space is a tensor, of type $(1, 0)$, which just means that it transforms the way elements of $V$ do (because it is an element of $V$) under change of coordinates for $V$.

Saying that something "is a tensor" doesn't tell you much until you specify what vector space it's a tensor relative to (and also ideally what its type is). This is complicated by the fact that physicists generally conflate tensors and tensor fields.

As EuYu says in the comments, it is also possible to sometimes interpret functions as being tensors of nontrivial type relative to other vector spaces.

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For any vector space $V$, you may consider the tensor algebra $\bigoplus_{n=0}^\infty V^{\otimes n}=V^{0}\oplus V \oplus (V\otimes V)\oplus\dotsb$

As a matter of convention, one usually sets $V^0$ to be the scalar field. Hence we would say the scalars live in the tensor algebra as tensors of rank zero, the vectors have rank one, and the higher tensors are there too.

In terms of coordinate changes, a scalar is something that gets multiplied by the zeroth power of the change of basis matrix. A vector gets a single change of basis, and a higher rank tensor gets more.

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