How to Taylor expand this function, which is a function of vectors? Suppose that $a,b,c,d$ are vectors in $\mathbb{R}^2$, so $a_{\alpha}$ where $\alpha = 1,2$. We define a scalar product as $a \cdot b := a_1 b_2 - a_2 b_1$. Suppose also that we have a real-valued function $f(a,b,c)$ which is a function of some $3$ vectors.
What does the Taylor expansion of $f$ look like to first order? For example, I want to Taylor expand $f(a+b,c,d)$ around $(a, c, d)$, so $$f (a+b, c, d) = f(a,c,d) + \ldots $$  but I don't know how to write the remaining terms. Do I get a scalar product of $b$ with some term?
 A: There is something to confront here in your question, which has been pointed out already. The first thing is to see that your "scalar product" is not a scalar product: it's a skew symmetric bilinear form. More specifically, notice that $a \cdot a = 0  \hspace{4pt} \forall a$. What you've written is actually the magnitude of the vector-cross product of $a$ and $b$ (More precisely, given a linear map $T: \mathbb{R}^2 \to \mathbb{R}^3 $, $a \cdot b = \vert Ta \times Tb \vert$). This is not a scalar product.
We can relate your skew symmetric form to the usual Euclidean inner product $( \cdot, \cdot)_E $ by the formula
$$
a \cdot b = (a, Jb)_E
$$
Where $J$ is a matrix,
$$
J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}
$$
So that's what we're dealing with. Now let's look at a Taylor expansion of your function. Any sensible Taylor expansion should look like
$$
F(a + tb) = F(a) + (b \cdot DF(a)) t + \mathcal{O}(t^2)
$$
Where $DF(a)$ is some suitable notion of derivative at $a$.
From this we can see that
$$
b \cdot DF(a) = \frac{d}{dt} F(a + tb) \vert_{t=0}
$$
I think that formula should get at the crux of your question. However, there's more to notice here. We can write out the right side as the usual inner product with the gradient $ \nabla F $:
$$
\frac{d}{dt} F(a + tb) \vert_{t=0} = (b, \nabla F(a))_E
$$
and using our relation of the skew symmetric form to the euclidean inner product, we have
$$
b \cdot DF(a) = (b, J DF(a))_E
$$
These two are equal, and if they are equal for all $b$, this implies
$$
J DF(a) = \nabla F(a)
$$
$J$ is invertible, so we see that $ DF(a) = J^{-1} \nabla F(a)$. Summarizing,
$$
F(a+b) = F(a) + b \cdot (J^{-1} \nabla F(a)) + \mathcal{O}(t^2)
$$
A: $\def\va{{\bf a}}
\def\vb{{\bf b}}
\def\vc{{\bf c}} 
\def\vd{{\bf d}}
\def\e{\varepsilon}
\def\np{\odot}$We have
\begin{align*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \sum_i b_i\left.\frac{\partial f(\va+\vb,\vc,\vd)}{\partial b_i}\right|_{b_i=0} + \ldots \\
&= f(\va,\vc,\vd) + \sum_i b_i \frac{\partial f(\va,\vc,\vd)}{\partial a_i} + \ldots \\
&= f(\va,\vc,\vd) + (\vb,\nabla_\va f(\va,\vc,\vd)) + \ldots, 
\end{align*}
where $(,)$ is the usual inner product and where
$\nabla_\va f = \langle \partial f/\partial a_1,\partial f/\partial a_2\rangle$ is the gradient.
But
$(\langle a_1,a_2\rangle,\langle b_1,b_2\rangle)
= \langle a_1,a_2\rangle\np\langle -b_2,b_1\rangle$.
(To avoid confusion with the dot product we will write $\va\np\vb=a_1b_2-a_2b_1$.)
Thus,
\begin{align*}
f(\va+\vb,\vc,\vd) &= f(\va,\vc,\vd) + \vb\np \widetilde\nabla_\va f(\va,\vc,\vd) + \ldots, 
\end{align*}
where
$\widetilde\nabla_\va f = \langle -\partial f/\partial a_2,\partial f/\partial a_1\rangle$.
Addendum
As is discussed in detail in the other answers, the product $\va\np\vb$ is only a "scalar product" in the sense that it is a product of vectors returning a real value.
A: If have only a partial answer, hope this helps:
The difficulty here is definitely your "scalar product". I set this in quotation marks because it is not a scalar product (not even a semi-definite one) because it is skew-symmetric: $a\cdot b = -b \cdot a$
It is merely a skew-symmetric bilinear form, which does not define a metric on your space. Now, I do not know much about exotic topologies, but I am pretty sure you need at least a sense of distance (i.e., a metric) to define derivatives on any space.
This is not given by your scalar product and therefore the Taylor expansion is not defined.
Is it possible that your "scalar product" is independent from the geometry of the space? I.e., it is just a skew-symmetric bilinear form defined in $\mathbb{R}^2$?
Then, the Taylor expansion would be the usual one with the usual scalar product $\langle a, b \rangle = a_1 b_1 + a_2 b_2$.
Let me know if I am wrong with anything!
