Wedge product of a 2-form with a 1-form.

Question

(Wedge product of a 2-form with a 1-form).*

Let $\omega$ be a $2-$form and $\tau$ a $1-$ form on $\mathbb R^3$. If $X, Y, Z$ are vector fields on $M$, find an explicit formula for $(\omega ∧\tau )(X,Y,Z)$ in terms of the values of $\omega$ and $\tau$ on the vector fields $X,Y,Z.$

Let $x^1,x^2,x^3$ are the coordinate function.$$\omega=f_1dx^1\wedge dx^2+ f_2dx^2\wedge dx^3+ f_3dx^1\wedge dx^3$$ $$\tau=g_1x^1+g_2 x^2+g_3x^3$$ $\omega \wedge \tau=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2$ $$X=a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3}$$ $$Y=b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}$$ $$Z=c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3}$$

$$\omega \wedge \tau=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2(X,Y,Z)$$ $$=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})=f_1g_3dx^1\wedge dx^2 \wedge x^3(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})+f_2g_1dx^2\wedge dx^3 \wedge dx^1(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})+f_3g_2 dx^1\wedge dx^3\wedge x^2(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})=f_1g_3a^1b^2c^3+f_2g_1a^2b^3c^1+f_3g_2a^1b^3c^2$$

Am I correct?

Answer 1

That's a physicist's answer, bristling with subscripts and superscripts.

I'd write $$(\omega\wedge\tau)(X,Y,Z)=\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y) +\omega(Y,Z)\tau(X).$$

I think you've lost some terms along the way...

Answer 2

Since we have by definition of the wedge product: $$(f\wedge g)(v_1,v_2,...,v_{k+l})={1\over k!l!}\sum_{\sigma\in S_{k+l}} (sgn \sigma)f(v_{\sigma(1)},v_{\sigma(2)}...,v_{\sigma(k)})g(v_{\sigma(k+1)},...,v_{\sigma(k+l)})$$

There should be a factor of $$1\over2$$ in the answer by Angina Sang. $$(\omega\wedge\tau)(X,Y,Z)={1\over2} \{\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y) +\omega(Y,Z)\tau(X)\}$$ There will not be a factor of $1\over2$ because we have used (k,l) shuffles.So the correct answer is $$(\omega\wedge\tau)(X,Y,Z)= \{\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y) +\omega(Y,Z)\tau(X)\}$$