# Confusion regarding the use of horizontal padding of $(1,1)$-tensor indices: which is the "correct" interpretation?

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 $$$$\tag{1} V^* \otimes V \cong V\otimes V \cong V \otimes V^*$$$$ (raise the first index, then lower the second one). This is fine, but I can't reconcile this with the linear map interpretation of $$(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.

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 "$$T^i_j$$".
• 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. Apr 26, 2021 at 14:26