# Proof that the derivative of a second order tensor w.r.t. a second order tensor is a fourth order tensor

We know that, since a linear map $$T$$ from a vector space $$V$$ to a vector space $$W$$ can be represented by a matrix, and, the derivative of a vector function $$f:V \rightarrow W$$ at $$a \in V$$ is the linear map $$D$$ for which, $$\lim_{h \rightarrow 0} \frac{\left\lvert f(a+h)-f(a)-D h \right\rvert}{|h|}=0$$ where $$h \in V$$ then $$D$$ can be represented by matrix. $$\def\tsr#1#2#3{{{#1}^{#2}}_{#3}}$$ $$\def\cT{\mathcal{T}}$$ $$\def\ltsr#1#2#3{{{}^{#2}}_{#3}{#1}}$$ $$\def\qty#1{\left[ #1 \right]}$$ $$\def\dF{\mathbb{F}}$$

A mixed second-order tensor is a bilinear map $$\ltsr{T}{1}{1}: V^* \times V \rightarrow \dF$$ (where $$\mathbb{F}$$ is the field over which vector space $$V$$ is defined and $$V^*$$ is the dual space of $$V$$), and $$\ltsr{\cT(V)}{1}{1}$$ is the vector space of mixed second-order tensors. There is more than one candidate for a linear map $$D: \ltsr{\cT(V)}{1}{1} \rightarrow \ltsr{\cT(V)}{1}{1}$$. We can define a linear map as, $$\def\vb#1{\mathbf{#1}} \def\cC{\mathcal{C}} Y=D(X) \iff \ltsr{Y}{1}{1} :=\ltsr{\cC}{1}{2} \left(\ltsr{D}{1}{1} \otimes \ltsr{X}{1}{1}\right) \tag{I}\label{I}$$ ($$\ltsr{\cC}{i}{j}:\ltsr{\cT(V)}{r}{s}\rightarrow \ltsr{\cT(V)}{r-1}{s-1}$$ is the contraction operator that applies the canonical pairing of the $$i$$th $$V^*$$ factor and the $$j$$th $$V$$ factor). Another way of defining the linear map is, $$Y=D(X) \iff \ltsr{Y}{1}{1} := \ltsr{\cC}{2}{2}\left(\ltsr{\cC}{2}{3} \left(\ltsr{D}{2}{2} \otimes \ltsr{X}{1}{1}\right)\right) \tag{II}\label{II}$$ where $$\ltsr{D}{2}{2} \in \ltsr{\cT(V)}{2}{2}$$ the vector space of bilinear maps $$\ltsr{T}{2}{2}: V^* \times V^* \times V \times V \rightarrow \dF$$.

Using heuristics, it is clear that, since we want to relate all components $$\tsr{Y}{j}{i}$$ to all components $$\tsr{X}{k}{l}$$, ($$\ref{II}$$) is the only linear map that can represent the derivative of $$Y$$ w.r.t. $$X$$ (($$\ref{I}$$) only relates the components of $$Y$$ that have the same contravariant index as $$X$$). But is there a rigorous justification for defining the derivative this way?

• By derivative I guess you mean a connection, which however is NOT a tensor. Under a given coordinate system it can be represented by 'a tensor' though. Commented Apr 23 at 14:34

I don't understand what you're trying to do. If you want "the derivative of a second order tensor w.r.t. a second order tensor", just take both $$V$$ and $$W$$ in your first paragraph to be $${}^1_1\mathcal T(V)$$. Then $$D : {}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$$ is a linear map. End of story. You haven't specified $$f$$ so it doesn't make sense to talk about "candidates for $$D$$". Even if you do specify $$f$$, it still doesn't make sense to talk about "candidates" because $$D$$ is necessarily unique.
If what you want is to understand how a linear map $${}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$$ is a fourth-order tensor, then what you need is the following: for any vector spaces $$U, V, W$$ there are natural isomorphisms $$U\otimes\mathbb F \cong U,\quad U\cong (U^*)^*,$$$$(V\otimes W)^* \cong V^*\otimes W^*,\quad V\otimes W\cong W\otimes V,$$$$U\otimes(V\otimes W) \cong (U\otimes V)\otimes W \cong U\otimes V\otimes W,$$$$\mathrm{Hom}(U\otimes V, W) \cong \mathrm{Hom}(V, U^*\otimes W).$$ Now noting that naturally an element of $${}^1_1\mathcal T(V)$$ can be considered a linear map $$V\otimes V^* \to \mathbb F$$, a map $${}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$$ is an element of $$\mathrm{Hom}([V\otimes V^*]^*, [V\otimes V^*]^*)$$ and so applying the above isomorphisms $$\mathrm{Hom}([V\otimes V^*]^*, [V\otimes V^*]^*) \cong \mathrm{Hom}(V^*\otimes V^{**}, V^*\otimes V^{**}) \cong \mathrm{Hom}(V^*\otimes V, V^*\otimes V) \cong \mathrm{Hom}(V^*\otimes V\otimes V\otimes V^*, \mathbb F) \cong \mathrm{Hom}(V\otimes V\otimes V^*\otimes V^*, \mathbb F).$$ This final space is naturally the same thing as $${}^2_2\mathcal T(V)$$. There is a little bit of choice here, in that we could independently swap the two copies of $$V$$ or the two copies of $$V^*$$.
• Thanks for the feedback, I will endeavor to improve the wording of the question. In my mind, the derivative of a second order tensor $X$ w.r.t. $X$ is a fourth order tensor $D=\delta^j_i \delta^l_k \mathbf{e}_j \otimes \mathbf{e}_l \otimes e^j \otimes e^k$. By the way you are right , all I am trying to understand is whether the linear map ${}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$ is a fourth-order tensor. Commented Apr 23 at 17:36
• @TedBlack Why the deltas? In general it will just have some arbitrary coefficients $D^{jl}_{ik}$. To be as concrete as possible, given $D : {}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$, you just plugin basis vectors to get the coefficients:　$$D^{jl}_{ik} = D(e^j\otimes e_i)(e^l, e_k)$$ where by $e^j\otimes e_i$ I mean the function $(v, \omega) \mapsto e^j(v)\omega(e_i)$ for all $v \in V$ and $\omega \in V^*$. Commented Apr 23 at 18:07
• The deltas are only for the specific case of the derivative of a second order tensor w.r.t. itself ($\partial X / \partial X$). Yes in general it would have some arbitrary coefficients ${D^{jl}}_{ik}$. Commented Apr 23 at 19:19
• You write "${}^1_1\mathcal T(V)$ can be considered a linear map $V\otimes V^* \to \mathbb F$, a map ${}^1_1\mathcal T(V) \to {}^1_1\mathcal T(V)$ is an element of $\mathrm{Hom}([V\otimes V^*]^*, [V\otimes V^*]^*)$". How did $\mathrm{Hom}(V\otimes V^*, V\otimes V^*)$ become $\mathrm{Hom}([V\otimes V^*]^*, [V\otimes V^*]^*)$? Commented Apr 24 at 10:55
• @TedBlack It didn't. A linear map $V\otimes V^* \to \mathbb F$ is by definition an element of $[V\otimes V^*]^*$, so ${}^1_1\mathcal T(V) \cong [V\otimes V^*]^*$. I'm trying to go step-by-step; saying that ${}^1_1\mathcal T(V) \cong V\otimes V^*$ requires applying the isomorphisms I listed in my answer, which is what I'm doing in the display equation after saying "and so appying the above isomorphisms". Commented Apr 24 at 14:21