TITLE: Second derivative of a vector field
QUESTION [9 upvotes]: I wonder how to treat the "second derivative" of a vector field. For example, imagine we have a vector field $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$. Then we evaluate the derivative at two points $Df(a)$ and $Df(b)$ which are matrices! Now,
$$D[Df(a)Df(b)] = D^2f(a)Df(b)+Df(a)D^2f(b).$$
My question is, what is $D^2f(a)$? How can I treat this? I imagine is something identifiable with $\mathbb{R}^{n\times n \times n}$. In such a case, if I wish to compute the "matrix" norm of $D[Df(a)Df(b)]$ (as the sum of all entries) is this then the sum of all possible combinations of
$$\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j} \frac{\partial}{\partial x_k} f(a)  \ ?$$
Thank you very much for your help!

REPLY [3 votes]: As already answered by SAUVIK, if $f:E\longrightarrow F$,
$$Df(x)\in L(E,F)\text{  and  } Df:E\longrightarrow L(E,F).$$
So,
$$D^2 f:E\longrightarrow L(E,L(E,F)).$$
Now, the space $L(E,L(E,F))$ can be identified of the space of bilinear functions form $E$ to $F$ via the isomorphism
$$g\to\hat g,\qquad\hat g(x,y) = (g(x))(y).$$
The trick can be obviously extended to higher orders.