Question about cross product of images of linear transformation

Question

I'm reading "Differential Geometry: Curves and Surfaces" of Manfredo Carmo, and this part in the book confuses me(page 166):

Suppose that $N: S \rightarrow S^2$ is the Gauss map of regular surface $S$ into unit sphere. We know that for each $p \in S$, $dN_p$ is a self-adjoint linear mapping. Suppose that $\{w_1, w_2\}$ is a basis in $T_p(S)$, then we have: $$dN_p(w_1) \times dN_p(w_2) = \det(dN_p)(w_1 \times w_2) = Kw_2 \times w_2$$

So the part confuses me is the equation $$dN_p(w_1) \times dN_p(w_2) = \det(dN_p)(w_1 \times w_2)$$

I did search around, and found this equation for cross product: $$Ma \times Mb = \det(M)(M^T)^{-1} (a \times b)$$

So, to make the mentioned equation right, we need to have $(M^T)^{-1} = I$, which I don't think it's true in general. Can anyone help me explain this? I really appreciate.

if any reader finds a reference for that formula, it would be helpful — user1868607, Apr 10 '18 at 16:57
There is something relevant for orthogonal (not self-adjoint) linear maps, but it applies only to maps on all of $\Bbb R^3$. See this post for a correct solution. I've also explained the cross-product formula you found. — Ted Shifrin, Aug 08 '20 at 17:16

score 0 · Accepted Answer · answered Feb 16 '16 at 04:15

0

The condition you've arrived at is exactly the "self-adjoint linear mapping" condition in the passage.

answered Feb 16 '16 at 04:15

Muphrid

19,902

Oh, I see. Seem like my formula is wrong. It should be $Ma \times Mb = \det(M)(M^T)^{-1}(a \times b)$. Thanks – le duc quang Feb 16 '16 at 05:45
This answer is total nonsense. – Ted Shifrin Aug 08 '20 at 17:15

Question about cross product of images of linear transformation

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