Relation between trilinear forms, Jacobi identity and closed $2$-forms

Question

Let $f : \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a trilinear mapping such that \begin{equation} f(x,y,z)=-f(x,z,y) \end{equation} and \begin{equation} f(x,y,z)+f(y,z,x)+f(z,x,y)=0 \end{equation}

From the first identity, I think $x \to f(x,\cdot,\cdot)$ becomes a $2$-form.

Next, it seems quite plausible from the second identity that the $2-$form $x \to f(x,\cdot,\cdot)$ is "closed".

However, I am quite confused about how to apply the exterior derivative to this $x \to f(x,\cdot,\cdot)$ and check its closedness..

Could anyone please clarify for me?

Answer 1

The idea of calculating $d\omega$ for $\omega$ the form $x\mapsto f(x,\cdot,\cdot)$ becomes somewhat simpler in coordinates.

Writing $f(x,y,z)=\sum_{\alpha,\beta,\gamma}f_{\alpha\beta\gamma}x^\alpha y^\beta z^\gamma$, where each $f_{\alpha\beta\gamma}$ is constant, we have the folowing rules $f_{\alpha\beta\gamma}+f_{\alpha\gamma\beta}=0$ and $f_{\alpha\beta\gamma}+f_{\beta\gamma\alpha}+f_{\gamma\alpha\beta}=0$. Then $\omega=\sum_{\alpha,\beta<\gamma}\frac{1}{2}(f_{\alpha\beta\gamma}-f_{\alpha\gamma\beta})x^\alpha dx^\beta\wedge dx^\gamma$. This gives \begin{align}d\omega&=\sum_{\alpha, \beta<\gamma} \frac{1}{2}(f_{\alpha\beta\gamma}-f_{\alpha\gamma\beta})dx^\alpha\wedge dx^\beta\wedge dx^\gamma\\ &=\sum_{\alpha<\beta<\gamma}\frac{1}{2}(f_{\alpha\beta\gamma}-f_{\beta\alpha\gamma}+f_{\beta\gamma\alpha}-f_{\alpha\gamma\beta}+f_{\gamma\alpha\beta}-f_{\gamma\beta\alpha})dx^\alpha\wedge dx^\beta\wedge dx^\gamma\\ &=\sum_{\alpha<\beta<\gamma}\frac{1}{2}(f_{\alpha\beta\gamma}+f_{\beta\gamma\alpha}+f_{\gamma\alpha\beta}-f_{\alpha\gamma\beta}-f_{\beta\alpha\gamma}-f_{\gamma\beta\alpha})dx^\alpha\wedge dx^\beta\wedge dx^\gamma\\ &=\sum_{\alpha<\beta<\gamma}(f_{\alpha\beta\gamma}+f_{\beta\gamma\alpha}+f_{\gamma\alpha\beta})dx^\alpha\wedge dx^\beta\wedge dx^\gamma=0\end{align}

EDIT: We may write that $f(x,y,z)=\sum f_{\alpha\beta\gamma}x^\alpha y^\beta z^\gamma$ because any trilinear map is determined by its value on each of the basis elements of $\mathbb{R}^n\otimes \mathbb{R}^n\otimes\mathbb{R}^n$ which are expressed as $e_\alpha\otimes e_\beta\otimes e_\gamma$

Answer 2

The relevant formula is the "conceptual" formulation of the exterior derivative. If $\omega$ is a $2$-form and $X,Y,Z$ are vector fields (here, on $\Bbb R^3$, we have \begin{multline*} d\omega(X,Y,Z) = X(\omega(Y,Z))+Y(\omega(Z,X))+Z(\omega(X,Y)) \\ - \omega([X,Y],Z) + \omega([X,Z],Y) - \omega([Y,Z],X). \end{multline*} In your case, we take $X,Y,Z$ to be the constant vector fields $e_i=\dfrac{\partial}{\partial x_i}$, $e_j=\dfrac{\partial}{\partial x_j}$, $e_k=\dfrac{\partial}{\partial x_k}$ on $\Bbb R^n$, and then $\omega(x)(e_j,e_k) = f(x,e_j,e_k)=a_{jk}(x)$. Here is the crucial point I missed at first: Since $f$ is trilinear, the functions $a_{jk}$ are linear functions of $x$, so $\frac{\partial f(x,e_j,e_k)}{\partial x_i}= f(e_i,e_j,e_k)$. The bracket terms all vanish because the vector fields commute. So we're left with \begin{align*} d\omega(e_i,e_j,e_k) &= e_i(f(x,e_j,e_k)) + e_j(f(x,e_k,e_i)) + e_k(f(x,e_i,e_j)) \\ &= f(e_i,e_j,e_k) + f(e_j,e_k,e_i) + f(e_k,e_i,e_j) = 0, \end{align*} as desired.

Answer 3

Let $\omega$ be the $2$-form corresponding to $f$. Let $u, v, w$ be three constant vector fields. The exterior derivative of $\omega$ in Euclidean space is given by the invariant formula

\begin{equation} (\mathrm{d}\omega)(u,v,w) = \mathcal{L}_u \omega(v, w) + \mathcal{L}_v \omega (w, u) + \mathcal{L}_w \omega(u, v). \end{equation}

Since $\omega$ is linear (as a function of the base point), we may write $$\mathcal{L}_u \omega(v, w) =\omega_u(v, w) - \omega_0(v, w) = f(u,v,w),$$ and similarly for the other terms. Hence $f(u,v,w) + f(v,w,u) + f(w,v,u) = 0$.