EDIT: I corrected some typos in my solution and wrote some details more precise (I was a bit sloppy first as I originally did not plan to discuss the full solution, but rather some ideas whether a solution can be constructed somewhat more naturally. E.g. using the Lie algebra and Lie group operations for $\mathbb{S}^3$.)

I am trying to solve the equation $$ \mathrm{d}G(x,y) = \mathrm{Vol}(x)+(-1)^{n}\mathrm{Vol}(y):= H $$ for $G\in \Omega^{n-1}\bigl((\mathbb{S}^n\times \mathbb{S}^n)\backslash \Delta=:M\bigr)$. Here $\mathrm{Vol}$ is the standard volume form on $\mathbb{S}^n$ and $\Delta$ the diagonal. The solution for $n=1$ is $$ G(x,y)= - \alpha(x,y)\quad \text{for all }x,y\in \mathbb{S}^1 $$ where $\alpha(x,y)$ is the counterclockwise angle from $x$ to $y$. I can find a solution for general $n\in \mathbb{N}$ as $$ G = \int_{[0,1]} \psi^*H $$ for a contraction $\psi: [0,1]\times N \rightarrow N$ of an open thickening $N$ of $M$ onto the antidiagonal $\overline{\Delta}:=\{(x,-x)\}$ (on $\overline{\Delta}$ holds namely $H=0$). Here I use a closed extension of $\mathrm{Vol}$ to $\mathbb{R}^{n+1}_{\neq 0}$. However, the solutions I obtain (in coordinates $x^i$ on $\mathbb{R}^{n+1}$) look too complicated and it is computationally hard to work with it further.

Does anybody have any idea how to solve this equation "nicely" e.g. by some natural construction of $G(x,y)$ similar to the case $n=1$?

I am mainly interested in the case $n=3$. Can the Lie group structure be used to construct a solution?

Thanks for any ideas!

Big picture: $H$ is a smooth integral kernel of the orthogonal projection to harmonic forms and $G$ is a (singular) integral kernel of the Green operator (:=the inverse of $\mathrm{d}$)

UPDATE1: As Robert Bryant suggests, the sign is indeed $(-1)^n$.

UPDATE2: Reading the nice answer of Robert Bryant, I just typed my complicated solution for a comparison. Perhaps it is the same except Robert using a clever notation: For every $n$ a solution is

$$ G(x,y)=\sum_{k=0}^{n-1} g_k(x\cdot y) \omega_k(x,y) $$ where $$ g_k(u) = \int_{0}^1 \frac{t^k(t-1)^{n-1-k}}{(2t(t-1)(1+u)+1)^{\frac{n+1}{2}}}\mathrm{dt} $$ can be solved inductively and $$ \omega_k(x,y) = \frac{1}{k!(n-1-k)!} \sum_{\sigma\in \mathbb{S}_{n+1}} (-1)^\sigma x_{\sigma_1}y_{\sigma_2}\mathrm{d}x_{\sigma_3}\ldots\mathrm{d}x_{\sigma_{2+k}}\mathrm{d}y_{\sigma_{3+k}}\ldots\mathrm{d}y_{\sigma_{n+1}}$$

P.S. I am still tempted to say something like $G(x,y)((v_1,w_1),\ldots,(v_{n-1},w_{n-1}))$ is the angle between the hyperplane $\langle w_1,\ldots,w_{n-1}\rangle$ at $y$ and the parallel transport of the hyperplane $\langle v_1,\ldots,v_{n-1}\rangle$ from $x$ to $y$ along the shortest arc on the unique great circle through $x$ and $y$. This might be a complete nonsense as well... If my computations are correct then for $n=2$ holds $$ G(e_1,\cos(\alpha)e_1 + \sin(\alpha)e_2) = \frac{\sin(\alpha)}{\cos(\alpha)-1}(\mathrm{d}y_3 - \mathrm{d} x_3 ). $$

  • 1
    $\begingroup$ Your final sentence in your P.S. above cannot be correct because, when $n=2$, $G$ is a $1$-form, so at a point $(u,v)\in M = \mathbb{S}^2\times\mathbb{S}^2\setminus\Delta$, it must be a linear map $$G_{(u,v)}:T_u\mathbb{S}^2\oplus T_v\mathbb{S}^2\to\mathbb{R},$$ which your proposed formula is not. [This comment was made before your most recent edit, where you removed your originally proposed formula for $G$ when $n=2$.] $\endgroup$ – Robert Bryant Jan 27 '18 at 15:33
  • 1
    $\begingroup$ The formula when $n=2$ that I got was $$G(x,y)(v,w) = \frac{(x\times y)\cdot (v-w)}{1-x\cdot y},$$ where $v\in T_x\mathbb{S}^2$ and $w\in T_y\mathbb{S}^2$. This agrees with your formula if, by $\mathrm{d}x_3$ and $\mathrm{d}y_3$ you mean the components of the vectors $v$ and $w$ in the direction $e_3 = e_1\times e_2$ in your notation. $\endgroup$ – Robert Bryant Jan 27 '18 at 15:40
  • $\begingroup$ Thanks! It seems to restrict to my formula in the particular case. $\endgroup$ – Pavel Jan 27 '18 at 15:43

NB: I have edited my answer to remove the comment at the beginning about correcting the original sign of the OP in the formula for $H$, since the OP has now corrected the erroneous sign in the question. I have also added one remark at the end about the case when $n$ is even.

Now, one has $$ H = x^*\Omega + (-1)^n\,y^*\Omega \tag1 $$ where $x:\mathbb{S}^n\times \mathbb{S}^n \to \mathbb{S}^n\subset \mathbb{R}^{n+1}$ and $y:\mathbb{S}^n\times \mathbb{S}^n \to \mathbb{S}^n \subset \mathbb{R}^{n+1}$ are the projections on the first and second factors and $\Omega$ is the $\mathrm{SO}(n{+}1)$-invariant volume form on $\mathbb{S}^n$ corresponding to a fixed orientation and inner product on $\mathbb{R}^{n+1}$.

Thus, to remove any sign ambiguities, fix an orientation and inner product on $\mathbb{R}^{n+1}$, and let $(u_0,\ldots,u_n)$ be an oriented orthonormal basis of $\mathbb{R}^{n+1}$, so that $U = u_0\wedge\cdots\wedge u_n$ is a basis of $\Lambda^{n+1}(\mathbb{R}^{n+1})$. Thus, for example, one has $$ \tfrac1{n!}\,x\wedge (\mathrm{d}x)^n = x^*\Omega\,U \quad\text{and}\quad \tfrac1{n!}\,y\wedge (\mathrm{d}y)^n = y^*\Omega\,U \tag2 $$ To save writing, for a form $\phi$ with values in $\Lambda^{n+1}(\mathbb{R}^{n+1})$, I will henceforth write $[\phi]$ for the scalar-valued differential form such that $\phi = [\phi]\, U$. Thus, $x^*\Omega = [\tfrac1{n!}\,x\wedge (\mathrm{d}x)^n]$, etc.

Using this notation, for example, one has, when $n=1$, that $\mathrm{d}G = H$ where $$ G = [\,f(x{\cdot}y)\,x\wedge y\,],\tag3 $$ where $f:[-1,1)\to\mathbb{R}$ is the (unique) smooth function that satisfies $$ f(\cos t) = \frac{(\pi-t)}{\sin t}\quad\text{for}\quad 0<t<\pi, $$ as one easily verifies by parametrizing as $x =(\cos\theta,\sin\theta)$ and $y=(\cos\phi,\sin\phi)$. Note that, since $x{\cdot}y<1$ on $M$, $G$ is a smooth function on $M$ and that $G$ vanishes when pulled back to the anti-diagonal, where $x+y=0$.

When $n=2$, calculation yields that $$ G = \frac{[x\wedge y\wedge \mathrm{d}(x{-}y)]}{1-x{\cdot}y}\tag4 $$ satisfies $\mathrm{d}G=H$, and, when $n=3$ (the case that the OP is particularly interested in), one finds that $$ G = \left[\,x\wedge y \wedge \bigl(\,f(x{\cdot}y)\,(\mathrm{d}x)^2 -\left(\tfrac12 + (x{\cdot}y)\,f(x{\cdot}y)\right)\mathrm{d}x\wedge\mathrm{d}y +f(x{\cdot}y)\,(\mathrm{d}y)^2\bigr)\,\right]\tag5 $$ satisfies $\mathrm{d}G = H$, where $f:[-1,1)\to\mathbb{R}$ is the (unique) smooth function that satisfies $$ f(\cos t) = \frac{\pi-t+\sin t\cos t}{4\sin^3t} \quad\text{for}\quad 0<t<\pi.\tag6 $$ (It may not be apparent that $f$ is smooth at $-1$, but it is, once one sets $f(-1) = \tfrac16$.)

The general pattern is clear, i.e., that, for general $n$, there will be an expression for $G$ of the form $$ G = \left[x\wedge y\wedge \left(\,f_0(x{\cdot}y)\,(\mathrm{d}x)^{n-1} + f_1(x{\cdot}y)\,(\mathrm{d}x)^{n-2}{\wedge}\mathrm{d}y + \cdots + f_{n-1}(x{\cdot}y)\,(\mathrm{d}y)^{n-1}\,\right)\right]\tag7 $$ where the $f_i$ are smooth functions on $[-1,1)$ that are determined by the condition that $\mathrm{d}G = H$. These coefficients can be determined recursively, but the general answer does not seem to be particularly illuminating, so I won't include it here.

However, I will remark that when $n$ is even, the functions $f_i$ turn out to be rational functions of $x{\cdot}y$. For example, when $n=4$, one finds $$ G = \frac{\left[\,x\wedge y\wedge\bigl(\,(\mathrm{d}x-\mathrm{d}y)^3 +(1-x{\cdot}y)\,((\mathrm{d}x)^3-(\mathrm{d}y)^3)\bigr)\,\right]} {18\,(1-x{\cdot}y)^2} $$


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