Is this actually a valid proof? Multivariable Calculus

Question

So I was asked to prove that the curl of a vector field $\mathbf F=(f_1,f_2,f_3)$ is invariant under change of basis, where the initial and final basis are both orthonormal. In particular, let the standard basis of $\mathbb R^3$ be $\mathbf e_1,\mathbf e_2$ and $\mathbf e_3$. Now for an orthogonal matrix $A=(a_{ij})$, construct a new basis as follows: $$\tilde{\mathbf e}_j=\sum_{i=1}^3a_{ij}\mathbf e_i$$ or $\tilde{\mathbf e}=\mathbf eA$ with the matrix notation.

Now I know that $\tilde D_j=\displaystyle\sum_i a_{ij}D_i$, which is basically the chain rule. Also, if we express $\mathbf F$ with our new basis as $\displaystyle\sum_j\tilde f_j\tilde{\mathbf e}_j$, then $f_i=\displaystyle\sum_ja_{ij}\tilde f_j$. Provided this, I'm supposed to prove $$\nabla\times\mathbf F=\sum_{i,j}(D_if_j)\mathbf e_i\times\mathbf e_j=\sum_{i,j}\left(\tilde D_i\tilde f_j\right)\tilde{\mathbf e}_i\times\tilde{\mathbf e}_j$$ A solution I have is as follows:

\begin{align*} \sum_{i,j}\left(D_if_j\right)\mathbf e_i\times\mathbf e_j&=(D_1\mathbf e_1+D_2\mathbf e_2+D_3\mathbf e_3)\times(f_1\mathbf e_1+f_2\mathbf e_2+f_3\mathbf e_3) \\ &=\mathbf e\nabla^t\times\mathbf e f \\ &=\mathbf e(A^{-1})^t\tilde\nabla^t\times\mathbf eA\tilde f \\ &=\mathbf eA\tilde\nabla^t\times\mathbf eA\tilde f \\ &=\mathbf e\tilde\nabla^t\times\mathbf e\tilde f \\ &=(\tilde D_1\tilde{\mathbf e}_1+\tilde D_2\tilde{\mathbf e}_2+\tilde D_3\tilde {\mathbf e}_3)\times(\tilde f_1\tilde{\mathbf e}_1+\tilde f_2\tilde{\mathbf e}_2+\tilde f_3\tilde{\mathbf e}_3) \\ &=\sum_{i,j}\left(\tilde D_i\tilde f_j\right)\tilde{\mathbf e}_i\times\tilde{\mathbf e}_j \end{align*} where $f$ is the vertical vector $(f_1,f_2,f_3)$.

Now what bugs me most is the first and second equality. Is it really OK to write it down like that, or are there underlying logic or calculations beneath it? I'm a freshman at college and not used to these kinds of notations. I tried to calculate this all by hand but was very unsuccessful. Any kind of help would be appreciated.

Answer 1

I think your argument is correct.

In the first equality, what you are actually doing is $$ \sum_{i,j}\left(D_if_j\right)\mathbf e_i\times\mathbf e_j =\sum_{i,j}\left(D_i\right)\mathbf e_i\times f_j\mathbf e_j =\left(\sum_{i} D_i\mathbf e_i\right)\times \left(\sum_jf_j\mathbf e_j\right). $$ This works fine by the associativity and commutativity of the sum. Formally it requires a proof by induction.

Your second equality is just a short-hand notation for the previous term. Had you wanted to, you could have done the proof with the notation from the first line (but it is your way that let's you see what's going on).

Answer 2

Your calculation looks fine to me. I did it from scratch like this, using only the definitions and orthonormality of $A$. (Too long for a comment, but I think it might be worthwhile posting it.)

$\sum_{i,j}\left(D_if_j\right)\mathbf e_i\times\mathbf e_j=$

$ (D_1A^T\tilde{\mathbf e}_1+ D_2A^T\tilde{\mathbf e}_2+ D_3A^T\tilde {\mathbf e}_3)\times( f_1A^T\tilde{\mathbf e}_1+ f_2A^T\tilde{\mathbf e}_2+ f_3A^T\tilde{\mathbf e}_3)=$

$(\tilde D_1\tilde{\mathbf e}_1+\tilde D_2\tilde{\mathbf e}_2+\tilde D_3\tilde {\mathbf e}_3)\times(\left(\sum_ja_{1j}\tilde f_j\right)A^T\tilde{\mathbf e}_1+ (\left(\sum_ja_{2j}\tilde f_j\right)A^T\tilde{\mathbf e}_2+(\left(\sum_ja_{3j}\tilde f_j\right)A^T\tilde{\mathbf e}_3)=$

$(\tilde D_1\tilde{\mathbf e}_1+\tilde D_2\tilde{\mathbf e}_2+\tilde D_3\tilde {\mathbf e}_3)\times(\tilde f_1\tilde{\mathbf e}_1+\tilde f_2\tilde{\mathbf e}_2+\tilde f_3\tilde{\mathbf e}_3)=$

$\sum_{i,j}\left(\tilde D_i\tilde f_j\right)\tilde{\mathbf e}_i\times\tilde{\mathbf e}_j$

Answer 3

While your calculation seems to be correct, one should note that the curl of a given vector field ${\bf F}$ is a vector-valued function which can be defined without recurrence to a basis in ${\mathbb R}^3$. It follows that your calculations are in fact unnecessary.

The linked Wikipedia article explains the following: Given a point ${\bf p}$ in the domain of ${\bf F}$ one can "measure" ${\rm curl}({\bf F})({\bf p})$ by drawing tiny discs $D$ with center ${\bf p}$ and surface normal ${\bf n}$. One then has $${\rm curl}({\bf F})({\bf p})\cdot{\bf n}\approx{1\over{\rm area}(D)}\int_{\partial D}{\bf F}\cdot d{\bf x}\ ,$$ and a corresponding limit relation. Fact is that ${\rm curl}({\bf F})({\bf p})$ encodes the "local nonconservativity" of ${\bf F}$ at ${\bf p}$. If ${\bf F}$ is a conservative field then the integrals $\int_{\partial D}{\bf F}\cdot d{\bf x}$ are all $=0$.