Let $\sigma : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ such that
$\sigma(x,u) = (x-x^3/3+xy^2,y-y^3/3+yx^2,x^2-y^2)$. Question is to show that this is a conformal map. Our definition is the following:
Let $S_1, S_2$ be a pair of regular surfaces. A diffeomorphism $f : S_1 \rightarrow S_2$ is called conformal if $df$ preserves angles, that is, if
$∠(u, v) = ∠(df_p(u), df_p(v))$ for all $p \in S1$ and all $u, v \in T_pS_1$, where
$∠(df_p(u), df_p(v))=cos^{-1}(\frac{\langle df_p(u), df_p(v) \rangle}{\lvert df_p(u) \rvert \cdot \lvert df_p(v) \rvert})$
I calculated the differential of this map but it is represented as a $3×2$ matrix. How can take the inner product of such matrices? Can you help me with this? Thanks for any help.
Best Answer
It is enough that the Gram matrix, which is 2 by 2, be scalar, meaning a (scalar function) multiple of the identity matrix. As you say your differential map is 3 by 2, call it $A,$ you need to just multiply to get $$ G = A^T A $$ The entries of $G$ are just dot products of vectors (each with three entries)