Pullback of a linear map on a 2-form.

Question

I am having a bit of trouble understanding a homework question and was seeking some clarification. Note, I have edited this question after I worked a couple of things out.

Given a 2-form $v=dx_1 \wedge dx_4+dx_2 \wedge dx_5+dx_3 \wedge dx_6$, I want to find an injective linear map $B:R^3 -> R^6$ such that $B^*v=0$.

So I was wondering, does $B^*v=0$ mean $v(B(x),B(y))=(dx_1 \wedge dx_4+dx_2 \wedge dx_5+dx_3 \wedge dx_6)(T(B(x)),T(B(y)))=0$ where $x,y \in R^3$ and T indicates the tangential derivative?

Assuming so, then $T(B(x))=B_i(x) \frac{\partial}{\partial x_i}=\sum_k c_k B_i(x_k) \frac{\partial}{\partial x_i}$.

So $B^*v=(dx_1 \wedge dx_4+dx_2 \wedge dx_5+dx_3 \wedge dx_6)(\sum_k c_k B_i(x_k) \frac{\partial}{\partial x_i},\sum_n d_n B_i(v_n) \frac{\partial}{\partial x_i}) =\sum_k c_k B_1(x_k) \sum_n d_n B_4(v_n)-\sum_k c_k B_4(x_k)\sum_n d_n B_1(v_n) +4 similiar terms = \sum_k \sum_n c_k d_n (B_1(x_k) B_4(v_n)- B_4(x_k) B_1(v_n))+ 2similiarterms=0$

But this is only true if $B_1(x_k) B_4(v_n)- B_4(x_k) B_1(v_n)=0$. This condition also holds when 1 & 4 are exchanged with 2 & 5 and 3 & 6. Is this what the answer, i.e. $B^*v=0$ iff B satisfies these conditions or do I need a more explicit form? Furthermore, how do I prove that this map is injective?

Thanks in advance!

Answer 1

Let $y_{k}=x_{k}=x_{k+3}$, $k=1,2,3$

$$B^{*}v=dy_{1}\wedge dy_{1}+dy_{2}\wedge dy_{2}+dy_{3}\wedge dy_{3}=0$$

Now, there are a few remarks to be made. The image of $\mathbb{R}^{3}$ under the map is not $\mathbb{R}^{6}$ but this should be expected. The best we could hope for by such a linear map is that the image is a 3-dimensional subset of $\mathbb{R}^{6}$. The map will preserve distinctness but will not be onto $\mathbb{R}^{6}$.

Answer 2

In response to a comment you posed:

$$x_{j}=\sum_{i=1}^{4}a_{ji}y_{i},\;\;\;j=1,...,6$$ Pulling back the given form I obtained

$$B^{*}v=\sum_{k,i=1}^{4}(a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k})y_{i}\wedge y_{k}$$

To make this vanish we respect the antisymmetric nature of the wedge product. All we need is for $a_{1i}a_{4k}+a_{2i}a_{5k}+a_{3i}a_{6k}$ to be symmetric when we replace $(i,j)$ with $(j,i)$. That is with an antisymmetric two form $y_{i}\wedge y_{k}$ and a symmetric coefficient we will sum over an antisymmetric summand. Said sum will vanish.