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Fix finite-dimensional vector spaces $V,W$. My whole life I'm been content with viewing linear maps $T:V\to W$ as $(1,1)$-tensors, i.e. elements of $W\otimes V^*$: such a $T$ yields a bilinear map $B:W^* \times V \to K$ defined by $B(\omega, v) = \omega(T(v))$. (I'm aware that the $(1,1)$-tensor terminology is generally reserved for the special case $W = V$ but keeping them distinct helps me keep things clear in my head for this particular issue.) In fixed bases I would be content to write things in coordinates as $$T = T^i_j \partial_i \otimes dx^j$$ and then feeding a vector $v = \partial_k v^k$ into $T$ (I am writing the basis vectors $\partial_k$ as components of a row of vectors) we would get the expected $$ T(\partial_k v^k) = \partial_i T^i_j v^k \delta^j_k = \partial_i (T^i_j v^j),$$ i.e., the column (component) vector of $T(v)$ is the product of the matrix of $T$ with the component vector of $v$.

I've been trying to learn some physics and stumbled across peculiar horizontal padding phenomenons in the context of Lorentz transformations which I've never seen before in mathematics. In this context, take $W = V$ (and this space will probably turn out to be a tangent space to flat spacetime, for example). Then one comes across $(1,1)$-tensors labeled $\Lambda^\mu{}_\nu$ and $\Lambda_\nu{}^\mu$. Mathematically, I would assume that the former is a coordinate representation of an element of $V \otimes V^*$ and the latter that of $V^* \otimes V$. From the context, one generally seems happy to raise and lower indices at will using a fixed metric tensor $g$ and the natural isomorphism $V^* \cong V$ it yields, and from there I am almost certain that the physicists have in mind the relation $$\Lambda^\mu{}_\nu = g_{i \nu}g^{j\mu} \Lambda_j{}^i$$ which corresponds to the isomorphism chain \begin{equation}\tag{1} V^* \otimes V \cong V\otimes V \cong V \otimes V^*\end{equation} (raise the first index, then lower the second one). This is fine, but I can't reconcile this with the linear map interpretation of $(1,1)$-tensors. As a mathematician, my first intuition when seeing $\Lambda^\mu{}_\nu$ and $\Lambda_\nu{}^\mu$ is to think of them as a single linear map, one represented by the matrix $\Lambda_\nu^\mu$. And this corresponds neatly to the isomorphism $W^* \otimes V \cong V \otimes W^*$ given by braiding, that is, simply swapping the two tensor components; these both naturally correspond to the space of linear maps $V \to W$. But the braiding $V^* \otimes V \cong V \otimes V^*$, independent of any choice, is not the same thing as the isomorphism $(1)$ above, which depends crucially on the given metric. And indeed it does not appear that the $\Lambda^\mu{}_\nu$ and $\Lambda_\nu{}^\mu$ are meant to represent the same tensor in slightly different bases.

My questions are: have I made any mistake in the analysis above? And if so, is there no way to reconcile both of these notions? Thanks for reading.


Some relevant threads:

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    $\begingroup$ I can't really tell if you're question is the same or different, but does math.stackexchange.com/questions/73171/… help? $\endgroup$ Apr 23, 2021 at 17:29
  • $\begingroup$ @HansLundmark Thanks, that did help quite a lot to get my thoughts straight! Much appreciated. $\endgroup$ Apr 24, 2021 at 15:01

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I think I managed to clear my own confusion as a byproduct of formulating my problem as accurately as possible. Thanks to HansLundmark for referring me to Index notation for tensors: is the spacing important? as well.

When seeing objects written in coordinates as $T^i{}_j$ and $T_j{}^i$, I would immediately jump to the conclusion that these must represent the same tensor (linear map) after a simple permutation of bases. In other words, I would assume that $$T = T^i{}_j \partial_i \otimes dx^j = T_j{}^i dx^j \otimes \partial_i,$$ so that acting on test vector $v = \partial_k v^k$ we should compute the same result $$T(v) = T(\partial_kv^k) = \partial_iT^i{}_jv^j = \partial_iT_j{}^iv^j.$$ We thus see that the assumption that these both represent the same map is equivalent to the assumption that the components are "symmetric", i.e. $$T^i{}_j = T_j{}^i,$$ in which case it is harmless to write $T^i_j$ for the linear map represented by either of these coordinate representations.

But this assumption is misguided: in general, when given a particular metric $g$ one shall consider instead that $$T^i{}_j = g_{\nu j}g^{ i \mu}T_\mu{}^\nu,$$ so that we don't generally want $T^i{}_j$ and $T_j{}^i$ to represent the same map. In this case there should be no general hope (or desire) to reconcile both under the same appellation "$T^i_j$".

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    $\begingroup$ Let me just state something you probably already realized: If you start with a linear map $T \colon V \rightarrow W$ between finite dimensional inner product spaces, identify it with a $"(1,1)"$ tensor in $W \otimes V^{*}$ whose components are $T^i{}_{j}$, and use the metrics to transform it to $T_{i}{}^{j}$, then the components $T_{i}{}^{j}$ are precisely the components of the adjoint map $T^{*} \colon W \rightarrow V$. If $V = W$ and $T$ is self-adjoint then indeed both represent the same map but in general they really are different maps. $\endgroup$
    – levap
    Apr 26, 2021 at 14:26

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