# Exercise with the pullback of tensor

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

• How are you defining $v_1^*$ and $v_2^*$? You're using the standard inner product to identify $\Bbb R^3 \cong \Bbb R^{3*}$? – 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