Consistent Definition of a Tensor

Question

I have come across two definitions for tensors. The first is as a map from $q$ copies of a vector space and $p$ copies of its dual space \begin{equation} T:\bigotimes^{p}V^{*}\bigotimes^{q}V\rightarrow\mathbb{R} \end{equation} and in particular a vector (an element of $V$) is a $(1,0)$ tensor that maps from $V^{*}$ to $\mathbb{R}$.

The second definition that I see is in terms of how the components (tensors themselves being coordinate independent) of a certain object transform under a change of basis. Usually, a rank $(p,q)$ tensor $T$ is defined as an object whose components transform under a coordinate transformation $L$ as \begin{equation} T'^{\mu_{1}\mu_{2}...\mu_{p}}_{\nu_{1}\nu_{2}...\nu_{q}}=L^{\mu_{1}}_{\alpha_{1}}L^{\mu_{2}}_{\alpha_{2}}...L^{\mu_{p}}_{\alpha_{p}}(L^{-1})^{\beta_{1}}_{\nu_{1}}(L^{-1})^{\beta_{2}}_{\nu_{2}}...(L^{-1})^{\beta_{3}}_{\nu_{3}}T^{\alpha_{1}\alpha_{2}...\alpha_{3}}_{\beta_{1}\beta_{2}...\beta_{q}} \end{equation} where undofruantely I don't know how to get the superscripts and subscripts to not sit line up without the tensor LaTeX package.

I am struggling to come up with a consistent way to reconcile these two ideas. Surely any object that satisfies the first definition does not necessarily satisfy the second, since exactly what a tensor is in the second definition depends on what the relevant coordinate transformations are. There is also a habit of writing objects that as far as I can tell aren't tensors (in either definition) in a way that makes them look like tensors. For example a Lorentz transformation $L$ of a vector $v$ with components $v^{\mu}$ in the original basis and $v'^{\mu}$ in the new basis $v'^{\mu}=L^{\mu}_{\nu}v^{\nu}$ makes it look like the object $L$ is a $(1,1)$ tensor but is not (from the first definition because the tensor mapping is independent of basis and so when written in terms of a specific basis involves only that basis, whereas a LT goes from one basis to another and $v$ and $v'$ are really the same object, and according to the second because the components of the LTs themselves do not change under an LT (so perhaps there is an argument to be made that it is a rank-0 tensor in the second definition)).

It makes sense that any link between the two would involve group theory, perhaps once the group of coordinate transformations is known objects of the first type that have certain group properties get called tensors by the second definition, but I don't really know. And that still leaves the problem that under the second definition what exactly a tensor is depends on whatever group of coordinate transformations you define. It would be unsatisfying if these definitions aren't equivalent but perhaps that is the case.

On a related note, when writing a general vector (in a basis) $v=v^{i}e_{i}$ and a dual vector $u=u_{i}e^{i}$ it seems the position of the index on $e^{i}$ or $e_{i}$ is not related to the type of tensor but is really just to make the summation convention work. $e_{i}$ are not components of a (0,1) tensor for example. Is this a correct analysis?

Thank you.

Answer 1

Surely any object that satisfies the first definition does not necessarily satisfy the second, since exactly what a tensor is in the second definition depends on what the relevant coordinate transformations are.

The idea is that vectors, covectors, and tensors of higher valence which are defined at the level of a vector space and its dual without reference to a particular choice of basis. Once we choose a basis $\hat e_i$ with dual basis $\hat \epsilon^i$, a $(p,q)$-tensor $T$ can be expressed in terms of its components $$T^{\mu_1,\ldots,\mu_p}_{\nu_1,\ldots,\nu_q} \equiv T\big(\hat \epsilon^{\mu_1},\ldots,\hat \epsilon^{\mu_p},\hat e_{\nu_1},\ldots,\hat e_{\nu_2}\big)$$ If we perform a change of basis $\hat e \mapsto L\hat e$ and $\hat \epsilon \mapsto L^{-1} \hat \epsilon$, then the transformation rules for the tensor's components follow immediately from via the tensor's multilinearity. So in that sense, an object which obeys the first definition has components which obey the second definition.

In the other direction, if an object has a list of $\mathbb R$-valued components which transform according to your second definition, then it can be "lifted" to the vector space level to define a well-defined basis-independent object.

There is also a habit of writing objects that as far as I can tell aren't tensors (in either definition) in a way that makes them look like tensors.

Index notation is used whenever there are a bunch of indices and sums to be performed. That's certainly true when working with tensors, but also with linear transformations. The context is important here.

$e_{i}$ are not components of a (0,1) tensor for example. Is this a correct analysis?

Yes, that's right. $\hat e_1, \hat e_2,\ldots$ are all vectors (i.e. (1,0)-tensors). $\hat e_i$ is the $i^{th}$ basis vector. Again, context is important when determining whether a particular symbol with an index (or indices) is a component of a tensor, a component of a linear transformation, or something else. The connection coefficients $\Gamma^i_{jk}$, for example, are not the components of a tensor.

I have a slight follow up question: applying this to a coordinate transformation such as a Lorentz transformation. I can see that under the first definition it is a tensor (it maps vectors to the same vector), but there are lots of answers on this site concerning whether LTs are tensors that say it is not a tensor because (for example) tensors are defined in terms of how they change under LTs, so LTs themselves are not tensors. I can however imagine taking an LT, transforming its components like a tensor (under an LT), and indeed I get the same components.

For a given finite-dimensional vector space $V$, there exists a canonical bijection between the set of $(1,1)$-tensors $V^*\times V \rightarrow \mathbb R$ and the set of linear transformations $V\rightarrow V$.

Letting $L$ be a linear transformation, we can define a $(1,1)$-tensor $T$ as $$T_L(\omega,X) := \omega\big(L(X)\big)$$ On the other hand, given a $(1,1)$-tensor $T$, we can define a linear transformation $$L_T(X) := T(\cdot, X)$$ where $\cdot$ denotes an open slot. The right-hand side is strictly speaking an element of $V^{**}$ which eats covectors and spits out real numbers, but for finite-dimensional $V$ there is a canonical isomorphism between $V$ and $V^{**}$.

An active Lorentz transformation is a linear map, not a tensor, but as per the above argument one could define a tensor from it. A passive Lorentz transformation is a change-of-basis. Such a thing acts on the lists of components, not on vectors directly, and so it is not actually a map from $V\rightarrow V$ ("vectors to vectors"), but rather a map from $\mathbb R^{\mathrm{dim}(V)} \rightarrow \mathbb R^{\mathrm{dim}(V)}$ ("components to components") though of course those spaces are isomorphic.