Let $V\subset\Bbb R^3$ be the subspace generated by vectors $v_1=(1,1,0)^t$,$v_2=(0,3,1)^t$, and let $\omega\in\Omega^2(\Bbb R^3)$ s.t. $\omega=2v_1^*\wedge v_2^*$.

I have the linear application $\phi:\Bbb R^2\rightarrow V$ s.t. $\phi (e_1)=v_1-v_2$ and $\phi (e_2)=v_1+v_2$, where $e_i$ is a vector of the canonical base; I need to find $\phi^t(\omega)$.

My solution goes like this: since {$v_1+v_2$, $v_1-v_2$} is a base too, for every $v\in V$ we have $v=av_1+bv_2=a'(v_1-v_2)+b'(v_1+v_2)=\phi(a'e_1+b'e_2)$, with $a'=\frac{a-b} 2,b'=\frac{a+b} 2$. So $\phi^t(\omega)=2(\frac 1 2e_1^*+\frac 1 2e_2^*)\wedge(-\frac 1 2e_1^*+\frac 1 2e_2^*)=e_1^*\wedge e_2^*$. However the solutions say that the result is $4e_1^*\wedge e_2^*$, so probably I'm wrong. Thank you in advance

  • $\begingroup$ How are you defining $v_1^*$ and $v_2^*$? You're using the standard inner product to identify $\Bbb R^3 \cong \Bbb R^{3*}$? $\endgroup$ – Ted Shifrin Jul 7 at 18:12

Note that the induced map $\phi_*$ sends $e_1\wedge e_2$ to $2v_1\wedge v_2$. This means that $\phi^*(v_1^*\wedge v_2^*) = 2e_1^*\wedge e_2^*$, and so $\phi^*(\omega) = \phi^*(2v_1^*\wedge v_2^*) = 4 e_1^*\wedge e_2^*$.

(Note that the matrix representation of $\phi$, using the standard basis for $\Bbb R^2$ and the given basis for $V$, is $\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$, and its determinant, $2$, is the factor that shows up in the computation. Of course, we can get it directly from $(v_1-v_2)\wedge (v_1+v_2) = 2v_1\wedge v_2$.)

REMARK: This computation is independent of how $v_1^*$ and $v_2^*$ are actually defined. :P


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