From what I understand of taking partial derivatives of tensor products, the following should be correct:
$\frac{\partial (m \otimes m)}{\partial m} = \frac{\partial m}{\partial m} \otimes m + m \otimes \frac{\partial m}{\partial m}$
$\frac{\partial (m \otimes m)}{\partial m} = I \otimes m + m \otimes I$
where m is a 6x1 vector and $m \otimes m$ is a 6x6 tensor.
My questions are:
- What would be the dimension of $I$? From the operation, it should be 6 x 1, but how can an identity tensor be a vector of size 6x1?
- If $I$ should just be treated like a numerical 1, the expression becomes $m + m$ which also doesn't look right.
$I$ is has dimensions "$6 \times 6$". In other words, $I \otimes m$ is a rank 3 tensor, with dimensions $6 \times 6 \times 6$. So $I$ is not literally to be treated as $1$, though it does act trivially on the space that $m$ lives in.
It might help to clarify with an example. Suppose that $m \in \mathbb{R}^6$ (I'm assuming you have some kind of vector space similar to Euclidean space if you're doing calculus in it).
Pick an orthonormal basis $\{e_i\}_{i=1}^6$ for $\mathbb{R}^6$. Then $I = \sum_{i=1}^6 e_i \otimes e_i$. So really you have $$ I \otimes m = \sum_{i=1}^6 e_i \otimes e_i \otimes m $$ which is a third order tensor. You can think of $(I \otimes m)_{ijk}$ as having three indices, each spanning from $1$ to $6$.
EDIT: Here's some more clarity on if you can simplify $I \otimes m + m \otimes I$. The summary is that you can't really simplify this expression any further, but if you expand $m$ in the orthonormal basis as $m = \sum_{i=1}^6 m_i e_i$ then you have $$\begin{align} \frac{\partial (m \otimes m)}{\partial m} &= I \otimes m + m \otimes I\\ &= \left(\sum_{i=1}^6 e_i \otimes e_i \otimes m \right) + \left(\sum_{i=1}^6 m \otimes e_i \otimes e_i \right)\\ &= \left(\sum_{i,j=1}^6 m_j \, e_i \otimes e_i \otimes e_j \right) + \left(\sum_{i,j=1}^6 m_j\, e_j \otimes e_i \otimes e_i \right)\\ &= \sum_{i,j=1}^6 m_j \left( e_i \otimes e_i \otimes e_j + e_j \otimes e_i \otimes e_i \right) \end{align}$$ Notice that $e_i \otimes e_i \otimes e_j + e_j \otimes e_i \otimes e_i$ can only be simplified when $i = j$, but if for example $i=1$ and $j=2$, then $$ e_1 \otimes e_1 \otimes e_2 \ne e_2 \otimes e_1 \otimes e_1 $$