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Given a differential form

$$x\,dy\wedge dz-y\,dx\wedge dz+z\,dx\wedge dy$$

I am supposed to prove that the it's pullback by a linear map of determinant one leaves it invariant. For example, if $\phi$ is a linear map and $\omega$ is a differential form then $\phi^*\omega=\omega$. Also, I was wondering, can we also say that any differential 2-form that is invariant under pull back of $\phi$ is scalar multiple of $\omega$?

I know the definition that for a linear map, we have that $\phi^*\omega(V)=\omega(\phi(V))$ but I don't know how to proceed.

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1 Answer 1

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Here's a hint: For any vectors $\mathbf v,\mathbf w\in\Bbb R^3$, $$\omega(\mathbf x)(\mathbf v,\mathbf w) = \det(\mathbf x,\mathbf v,\mathbf w).$$ Here I'm writing $\mathbf x=(x,y,z)$.

ADDENDUM: Here's a suggestion to address your uniqueness question. Consider two types of matrices with determinant $1$: $$\begin{bmatrix} c^2 & 0 & 0 \\ 0 & 1/c & 0 \\ 0 & 0 & 1/c \end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$$ and start with a $2$-form like $f dx\wedge dy$.

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  • $\begingroup$ thanks for you hint. Do you have hints for the later part of my question, any 2-form that satisfies the given invariance is scalar multiple of $\omega$ defined above. $\endgroup$
    – user147677
    May 4, 2014 at 13:01
  • $\begingroup$ See the addendum. Have fun! $\endgroup$ May 4, 2014 at 15:23
  • $\begingroup$ @TedShifrin: Sorry for necropost: By the notation det(x,v,w), do you mean each of x,v,w as column vectors? $\endgroup$
    – MSIS
    Jul 6, 2021 at 23:26
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    $\begingroup$ @MSIS: I do, but it doesn’t matter if you use row vectors for them all. $\endgroup$ Jul 7, 2021 at 0:54

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