4
$\begingroup$

From my understanding, a vector is a geometric object, which can be expressed as $$ v = v_i e^i $$ where $v_i$ and $e^i$ are components and basis, respectively.

It seems to me that many people, e.g. in wikipedia https://en.wikipedia.org/wiki/Tensor#Definition

regard tensor as its component, e.g. $$T_{i_1...i_n}^{j_1 ... j_n}$$

After all, is the tensor defined as the component, $$T_{i_1...i_n}^{j_1 ... j_n}$$ or including basis? $$T = T_{i_1...i_n}^{j_1 ... j_n} e^{i_1...i_n}_{j_1 ... j_n} $$

$\endgroup$
  • $\begingroup$ In physics a tensor is a physical quantity/property of an object that has certain transformation properties under coordinate transformations. Ontologically this goes beyond the definition by mathematics and assumes that one can assign a certain level of reality to tensors (one can, for instance, give experimental instructions on how to measure the components of a stress tensor). The basis is given by an (arbitrary or convenient) choice of coordinate vectors (which also have physical realizations), components would be the measurements relative to this basis. Neither is necessary, though. $\endgroup$ – CuriousOne Oct 28 '15 at 5:50
  • 2
    $\begingroup$ @CuriousOne I never liked the "physicist" notion of a tensor. I prefer to think of a tensor as a real (or complex) valued function of a prescribed number of vectors and covectors which is linear in every element. This gets your the transformation business but has the advantage that you remember that components are just representations of the underlying thing in a particular basis. I realize this isn't always the best way to think of a tensor, but in my experience it's usually the best way. $\endgroup$ – DanielSank Oct 28 '15 at 6:09
  • 1
    $\begingroup$ @DanielSank: Components are indeed not necessary. Misner, Thorne and Wheeler went out of their way to write an entire GR textbook both in coordinate free and component representation in parallel. You are the judge which way of formulating the same reality is better... I tried to work my way trough coordinate and component free physics (both in GR and on simpler problems), but I found that many times the advantages were marginal. When something is as simple as a scalar or vector product, it's all good, but as soon as one has to diagonalize a matrix and a computer is needed components are back. $\endgroup$ – CuriousOne Oct 28 '15 at 7:13
2
$\begingroup$

This isn't really a physics question, but while it stays up I can give a quick response:

Your are correct that a tensor is often defined in terms of its coordinate transformations, but this is not the only or most modern definition. Here is a coordinate-free definition:

A tensor of type (r,s) on a vector space $V$ and its dual $V^*$ is a complex-valued function on

$V_1\times V_2 \times ...V_r \times V^*_1 \times V^*_2 \times ... V^*_s$

(where $V_i$ are just numbered copies of the same vector space and likewise for $V^{*}_i$) that is linear in each argument.

Here $r+s$ is the rank of the tensor.

When you have an inner product, this formalizes the way that Einstein summation naturally suggests you think about tensors: a tensor of rank $n$ is an object that, when 'fed' $n$ vectors, spits out a scalar. The transformation properties can also be derived from this definition quite directly.

This definition is taken from the book An Introduction to Tensors and Group Theory for Physicists, which can offer much more elaboration on what this means.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer. Could you somehow transform your modern definition in the context of my question, i.e., is the definition for $T_i^j$ or $T_i^j e_j^i$ or something else? $\endgroup$ – Joe Oct 28 '15 at 6:21
  • $\begingroup$ This is the definition for $T$ itself, without any choice of coordinates. Given a particular basis, one can find the actions of $T$ on that basis and use that to construct the components $T_i^j$. $\endgroup$ – Rococo Oct 28 '15 at 23:46
  • $\begingroup$ Is the fact that a tensor is a 'complex-valued function' dependent on V being a complex vector space? In Lee's book /Introduction to Smooth Manifolds/, I've only seen tensors referred to as real multilinear maps (granted I've not read it cover to cover and my understanding is tenuous at best). $\endgroup$ – Paul Wintz Dec 8 '19 at 12:01
  • $\begingroup$ Well, you can certainly have a complex-valued tensor that acts on a real vector space. A trivial example is $T(x,y)=(x,iy)$, where the space is $\mathcal{R}^2$ (spanned by real numbers x and y). $\endgroup$ – Rococo Dec 12 '19 at 1:26
2
$\begingroup$

Given any $\mathbb{R}$-vector space $V$ and its dual space $V^\ast$, a tensor of rank $(k,l)$ is a map $$ T : V\times\dots V\times V^\ast\times\dots\times V^\ast \to \mathbb{R}\tag{1}$$ where $V$ occurs $k$ times and $V^\ast$ occurs $l$ times, that is linear in each argument. Equivalently, it is an element of the tensor product $V^\ast\otimes\dots\otimes V^\ast\otimes V\otimes\dots\otimes V$.

Now, if we choose a basis $e_i$ for $V$ and the corresponding dual basis $e^i$ of $V^\ast$ defined by $e^i(e_j) = \delta^i_j$, the components of the tensor are obtained as $$ {T_{i_1\dots i_k}}^{j_1\dots j_l} := T(e_{i_1},\dots,e_{i_k},e^{j_1},\dots,e^{j_l})\tag{2}$$ If we think of the tensor as an element of the tensor product, then the tensor product has a natural basis in the tensors $e^{i_1}\otimes\dots\otimes e^{i_k}\otimes e_{j_1}\otimes\dots\otimes e_{j_l}$ and the full tensor $T$ is described by $$ T = {T_{i_1\dots i_k}}^{j_1\dots j_l}\left(e^{i_1}\otimes\dots\otimes e^{i_k}\otimes e_{j_1}\otimes\dots\otimes e_{j_l}\right)\tag{3}$$ with a sum over all indices on the r.h.s. implied. That both of these descriptions of the tensor are equivalent follows directly from using the dual relations $e^i(e_j) = e_j(e^i) = \delta^i_j$.

It does not matter whether a tensor is defined "in components" as in $(2)$, as an abstract map as in $(1)$ or as the sum of basic tensors as in $(3)$. All these notions are equivalent.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.