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In some ways, this is a question about notation.

In my experience, I have only seen the classical Lie groups — such as $\operatorname{GL}(n,\mathbb{R})$, $\operatorname{SL}(n,\mathbb{R})$, $\operatorname{O}(n,\mathbb{R})$, $\operatorname{SO}(n,\mathbb{R})$, $\operatorname{Sp}(2n,\mathbb{R})$, etc. — introduced and defined in terms of (real) matrices (I'm sticking with $\mathbb{R}$ as the field just to keep things simpler).

My question: can a 2nd order tensor be a member of one of these groups? Or is it only the matrix representation of a tensor that can be a member of one of these groups? If the answer is the latter, then are there corresponding tensor groups for tensors whose matrix representation (in any basis) is a member of one of the classic Lie groups?

Example: The general linear group, $\operatorname{GL}(n,\mathbb{R})$, is defined for matrices simply as the group of all non-singular real square matrices: $$ \operatorname{GL}(n,\mathbb{R}) \;=\; \{ A\in M_{n,n}(\mathbb{R}) \;|\; \text{det}(A)\neq 0 \} $$

where $M_{n,n}(\mathbb{R})$ denotes $n\times n$ real matrices. Now, consider some finite n-dimensional (real) vector space, $V$, with an inner product. If we have some (1,1)-tensor $\mathbf{Q}\in \mathcal{T}^1_1(V)$ with non-zero determinant (i.e., $\mathbf{Q}$ is non-degenerate), then can we say that $\mathbf{Q}\in\operatorname{GL}(n,\mathbb{R})$? Or maybe the proper notation would be $\mathbf{Q}\in\operatorname{GL}(V)$?

Although I used a (1,1)-tensor for this example, I don't think anything would change if $\mathbf{Q}$ were a (2,0)-tensor or a (0,2)-tensor, would it? (I am assuming we have an inner product and can thus raise and lower indices as we please)

Does anything change for the other classical Lie groups? For instance, if $[Q]$ is the matrix representation (in any basis) of $\mathbf{Q}$ and satisfies $[Q][Q]^{\top}=I$ such that $[Q]\in\operatorname{SO}(n,\mathbb{R})$, can we say $\mathbf{Q}\in\operatorname{SO}(n,\mathbb{R})$? Or perhaps $\mathbf{Q}\in\operatorname{SO}(V)$? (this would mean that $\mathbf{Q}\cdot\mathbf{Q}^{\top}=\delta^i_j\mathbf{e}_i\otimes\boldsymbol{\epsilon}^j=\mathbf{I}$, where, here, the dot denotes a contraction using the inner product/metric, $\mathbf{e}_i$ and $\boldsymbol{\epsilon}^i$ are basis vectors and basis 1-forms for $V$ and $V^*$, and $\mathbf{I}$ is the identity tensor for $V$. )

Another question: same question for the associated Lie algebras. I.e., can a tensor be a member of $\mathfrak{so}(n,\mathbb{R})$ or only a matrix?


Edit: To clarify, I'm using the term tensor such that $T\in \mathcal{T}^p_q(V)$ means $T$ is (p+q)-linear map, $T:(V^*)^{\times p} \times V^{\times q} \to \mathbb{R}$. That is, $T= V^{\otimes p}\otimes (V^*)^{\otimes q}$. Under a basis transformation $\mathbf{e}'_i = M^j_{\;i}\mathbf{e}_j$ and $\boldsymbol{\epsilon}'^i = W^i_{\;j}\boldsymbol{\epsilon}^j$ (with $M^i_{\;j}W^j_{\;k}=\delta^i_k$), tensor components would then transform as
$$T' ^{i_1\dots i_p}_{\;\;\; j_1 \dots j_q} =( W^{i_1}_{\;k_1}\dots W^{i_p}_{\;k_p} M^{l_1}_{\;j_1}\dots M^{l_q}_{\;j_q} ) T ^{k_1\dots k_p}_{\quad l_1 \dots l_q}$$


Edit: I have little knowledge of group theory beyond the basic definition of a group and the various classical groups, which I only know as matrix groups. But I like tensors more than matrices for the work I typically do. Hence my question.

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  • $\begingroup$ “tensor” implies some properties under transformations. Here, what would be these transformations? $\endgroup$ Commented Jul 25, 2022 at 0:16
  • $\begingroup$ I'm using tensor such that $T\in \mathcal{T}^p_q(V)$ means $T$ is (p+q)-linear map, $T:(V^*)^{\times p} \times V^{\times q} \to \mathbb{R}$. That is, $T= V^{\otimes p}\otimes (V^*)^{\otimes q}$. Under a basis transformation $\mathbf{e}'_i = M^j_{\;i}\mathbf{e}_j$ and $\boldsymbol{\epsilon}^i = W^i_{\;j}\boldsymbol{\epsilon}^j$ (with $M^i_{\;j}W^j_{\;k}=\delta^i_k$), Tensor components would then transform as $T' ^{i_1\dots i_r}_{\;\;\; j_1 \dots j_s} =( W^{i_1}_{\;k_1}\dots W^{i_p}_{\;k_p} M^{l_1}_{\;j_1}\dots M^{l_q}_{\;j_q} ) T ^{k_1\dots k_p}_{\quad l_1 \dots l_q}$ $\endgroup$
    – J Peterson
    Commented Jul 25, 2022 at 1:49
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    $\begingroup$ representations by matrices are then not tensors as they satisfy $R(\omega_1)R(\omega_2)=R(\omega_1 \circ \omega_2)$. They are just matrices. $\endgroup$ Commented Jul 25, 2022 at 1:58
  • $\begingroup$ "representations by matrices are then not tensors" - Yes I know that a matrix is not a tensor but Im not sure what the last part of your comment means. But there are second-order tensors with properties analogs to the matrix Lie groups I mentioned. What I mean to ask is: if the matrix representation of a tensor (which itself is not a tensor) is a member of some Matrix lie group, what then can we say about the tensor itself? Is there some corresponding group the tensor belongs to? $\endgroup$
    – J Peterson
    Commented Jul 25, 2022 at 4:25
  • $\begingroup$ a tensor would transform multilinearly. Matrices (as group elements) do not. $\endgroup$ Commented Jul 25, 2022 at 5:04

2 Answers 2

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Any type of mathematical object in the right representation can belong to a group as long as the necessary requirements are met. Tensors may in fact be brought in a form as to belong to some group. A matrix representation of a tensor may potentially belong to a Lie group with matrices, or in an $n \times n$-tuple form, they may belong to an appropriate group consisted of such objects.

The key point here is that tensors, as you said, are multilinear maps from a number of copies of a certain vector space $V$ and its covector space $V^{*}$ to $\mathbb{R}$ (assuming that the target vector space is real). By contrast, in any group $G$ there is only one type of mathematical action:

\begin{equation} g_{1} \circ g_{2} = g_{3} \, , \quad g_{1}, g_{2}, g_{3} \in G \end{equation}

This means that tensors carry additional mathematical information which is not implicit to any group. What this translates to is that, if a tensor $\mathbf{T} ^{p}{}_{q}$ is in some representation $r$ in the form $T_{r}$ so that $T_{r} \in G$ (where $G$ some group), then:

\begin{equation} r: \mathbf{Q} ^{p}{}_{q} \rightarrow Q_{r} \, , \quad Q_{r} \neq T_{r} \not\Rightarrow Q_{r} \in G \end{equation}

In other words, just because two tensors may share the same characteristics (dimensionality, valence and target vector space), that doesn't mean that if one belongs to some group in a certain representation then the same representation of the second one also belongs to the same group. Similarly, just because a certain representation of some tensor belongs to a group, that doesn't mean any other random group element also satisfies the definition of a tensor.

So the answer to your first (couple of) question(s) is a reluctant kind of yes: a tensor may in fact be a member of a Lie group, including the classical ones if you can find a representation in which the tensor satisfies the requirements to belong to said group. The caveat is that it is only this specific form that belongs to it, and its membership to the group relies solely on its characteristics as an object in the representation it is in, not its tensor qualities. A matrix representation of a tensor might belong to a group, but this matrix we chose to represent it by no longer implies that it satisfies the definition of a tensor. For example, you may find an inertia tensor that in its matrix form is the identity matrix, and the identity matrix could belong to a variety of groups. However, this element when encountered as a member of the group doesn't imply any of the characteristics of a tensor. Essentially as I see it, this is tantamount to saying that it is a more stringent requirement to be a tensor rather than a member of some group.

As for the last question, there is a similar answer. Since Lie algebras are bilinear maps of vector spaces themselves, usually you can't have elements be tensors, as in to satisfy the same necessary requirements to be defined as such. You can always take a tensor and put it in a representation that could resemble an object of such a vector space of the algebra (e.g. in an $n$-tuple form), but since a vector space is not simply a group, due to satisfying additional mathematical requirements, the resulting element may cease to be a tensor. It would be an element of the vector space we are examining that simply happens to have the same (ordered) constituents as the tensor in said representation. An example of a case where tensors are in fact elements of a Lie algebra which I missed is what Qmechanic mentions below.

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  1. On one hand, a multiplicative Lie group $G$ is not a vector space, and hence cannot be a tensor space.

  2. On the other hand, a Lie algebra $\mathfrak{g}$ can sometimes be a tensor space.

    Example: The EM field strength $$F~\in~\bigwedge\!{}^2\mathbb{R}^{3,1}~\cong~ \mathfrak{so}(3,1)$$ in SR is a tensor under Lorentz transformations $\Lambda\in O(3,1)$.

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  • $\begingroup$ Regarding your first point, aren’t the linear/matrix Lie groups I mentioned also vector spaces? They meet all the vector space requirements regarding vector addition and scalar multiplication I believe. Isn’t their “group-ness”, closure under matrix multiplication, just additional structure on top of their “vector space-ness”? $\endgroup$
    – J Peterson
    Commented Jul 25, 2022 at 7:48
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    $\begingroup$ No, the $0$ matrix does not belong to a multiplicative matrix group. $\endgroup$
    – Qmechanic
    Commented Jul 25, 2022 at 7:52

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