# Give a formula for the pairing <α∧β,v⊗w> in terms of the pairings of vectors and 1-forms by expressing the 2-form as a tensor and contracting.

I am learning tensor and metric, and I found a question which I don't know how to solve. The question is:

Let $\alpha$ and $\beta$ be 1-forms, v and w vector fields. Give a formula for the pairing $<\alpha\wedge\beta,v\otimes w>$ in terms of the pairings of vectors and $1$-forms by expressing the $2$-form as a tensor and contracting.

I was guessing whether we could write $\alpha\wedge\beta$ as $\frac{1}{2}(\alpha\otimes\beta -\beta\otimes\alpha)$, since I know there exist these kinds of maps mapping 2-forms to type (0,2) tensors. However, does it mean $\alpha\wedge\beta =\frac{1}{2}(\alpha\otimes\beta -\beta\otimes\alpha)$? I really doubt.

Note:<,> refers to a contraction between a tangent vector and a 1-form, i.e. $w=\sum_{i}w_{i}dx^{i}$, $v=\sum_{j}v^{j}\dfrac{\partial}{\partial x^{j}}$, $<w,v>=\sum_{ij}w_{i}v^{j}<dx^{i},\dfrac{\partial}{\partial x^{j}}>=\sum_{ij}w_{i}v^{j}\delta_{j}^{i}$, where $\delta_{j}^{i}$ is the Kronecker Delta.

• The question should probably be "Give a formula for the pairing..." Jun 5, 2018 at 11:32
• You are right. Just corrected it. Thx! Jun 5, 2018 at 11:36
• What is the original definition of the pairing $\langle \alpha\wedge\beta,v\otimes w \rangle$? Jun 5, 2018 at 12:35
• I wrote it down in the bottom as NOTE. That is the only definition about the pairing I can find in the material. Jun 5, 2018 at 12:52
• This is the formula for the pairing of vectors and 1-forms, while you are supposed to provide a formula for the pairing of 2-forms and (2,0)-tensors. What is the definition of the latter pairing? Jun 5, 2018 at 13:07

If we dub $\alpha=a_sdx^s$ and $\beta=b_sdx^s$ and choose to define $$\alpha\wedge\beta=\alpha\otimes\beta-\beta\otimes\alpha$$ and with $v=v^{\mu}\partial_{\mu}$ and $w=w^{\nu}\partial_{\nu}$ then $$\alpha\wedge\beta(v,w)=\alpha\otimes\beta(v,w)-\beta\otimes\alpha(v,w)$$ $$=\alpha(v)\beta(w)-\beta(v)\alpha(w)$$ For which you evaluate 1st as $$\alpha\wedge\beta(\partial_i,\partial_j)=\alpha(\partial_i) \beta(\partial_j)-\beta(\partial_i)\alpha(\partial_j)$$ $$=a_ib_j-b_ia_j\qquad\qquad (1)$$ then $$\alpha\wedge\beta(v,w)=\alpha(v^i\partial_i)\beta(w^j\partial_j)-\beta(v^i\partial_i)\alpha(w^j\partial_j)$$ $$=v^iw^j\alpha(\partial_i)\beta(\partial_j)-v^iw^j\beta(\partial_i)\alpha(\partial_j)$$
$$=v^iw^j\alpha(\partial_i)\beta(\partial_j)-v^jw^i\beta(\partial_j)\alpha(\partial_i)$$
$$=(v^iw^j-v^jw^i)\beta(\partial_j)\alpha(\partial_i)$$
$$=(v^iw^j-v^jw^i)\alpha(\partial_i)\beta(\partial_j)$$ $$=(v^iw^j-v^jw^i)(a_ib_j-b_ia_j)$$ where we have subbed (1) above, still with the index-above index-below sum-convention.
• I think that this number is (exemplifying at dimension 3) $$\det\left(\begin{array}{cc}v^1&w^1\\v^2&w^2\end{array}\right)\det\left(\begin{array}{cc}a_1&b_1\\a_2&b_2\end{array}\right)+\det\left(\begin{array}{cc}v^2&w^2\\v^3&w^3\end{array}\right)\det\left(\begin{array}{cc}a_2&b_2\\a_3&b_3\end{array}\right)+\det\left(\begin{array}{cc}v^1&w^1\\v^3&w^3\end{array}\right)\det\left(\begin{array}{cc}a_1&b_1\\a_3&b_3\end{array}\right)$$ Jun 7, 2018 at 19:53