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Given a nondegenerate real cubic form $f(x,y)$ in two variables, consider the integral

$$I(f) = \frac{|\mathrm{Disc}(f)|^{1/6}}{2\pi} \int_0^{2\pi} f(\cos(\theta),\sin(\theta))^{-2/3} \, d\theta$$

The paper of Bhargava-Shankar-Tsimerman about Davenport-Heilbronn theorems asserts (pg. 468) that $I(f)$ does not depend on $f$ as we vary within the set $V_{\mathbf{R}}^{\pm}$ of nondegenerate real cubic forms whose discriminants $\mathrm{Disc}(f)$ have a fixed sign $\pm$.

These $V_{\mathbf{R}}^{\pm}$ also happen to be orbits of the natural action of $G = \mathrm{GL}_2(\mathbf{R})$ on the space of two-variable cubic forms. So equivalently they are asserting that $I(f)$ is invariant under the action of $G$, on these orbits.

I would like to understand why this invariance holds.

As a follow-up question, is this invariance a special case of some more general fact about integrals of (powers of) homogeneous polynomials around the unit circle being invariant?

$$$$ $$$$ $$$$

Addendum:

Thanks to Stanley Yao Xiao for a really helpful answer. I wanted to follow up the discussion a bit, to understand better this passage in Bhargava-Shankar-Tsimerman.

I am puzzled by the authors’ assertion on pg. 468 that "the ratio in (48) is independent of the $K$-invariant set $B$. Thus, …, (48) is equal to …"

Are we varying the set $B$?

Since $B$ is a subset of $V_{\mathbf R}$, the numerator and denominator of (48) are both integrals in $\mathbf R^4$, but $\displaystyle \int_K$ is a line integral, so why ``Thus”?

Is the following the correct argument?

Recall that $H^{(i)}$ is the largest subset of $\mathrm{GL}_2(\mathbf{R})$ such that $H^{(i)} \cdot v_i = B \cap V_{\mathbf R}^{(i)}$. Then by Proposition 23, (48) is equal to

$(*)$ \begin{equation} \frac {\displaystyle \frac{2\pi}{n_i} \int_{H^{(i)}} |\mathrm{Disc}(g \cdot v_i)|^{1/6} |a(g\cdot v_i)|^{-2/3} dg} {\displaystyle \frac{2\pi}{n_i} \int_{H^{(i)}} dg} \label{xx} \end{equation} Write $\mathrm{GL}_2(\mathbf{R}) = KL$ where $K=\mathrm{SO}(2)$, $L=AN\Lambda$. Then $dg=dk \, dl$ where $dk$ is the Haar probability measure on $K$ and $dl$ is the right Haar measure on $L$. Since $B \cap V_{\mathbf R}^{(i)}$ is invariant under $K$, it follows that $H^{(i)}$ is invariant under left multiplication by $K$, and thus we have $H^{(i)} = KL^{(i)}$ for some subgroup $L^{(i)} \subseteq L$. Then $(*)$ is equal to

$(**)$ \begin{equation} \frac{\displaystyle\int_{K \times L^{(i)}} |\mathrm{Disc}(kl\cdot v_i)|^{1/6} |a(kl \cdot v_i)|^{-2/3} dk \, dl} {\displaystyle\int_{K \times L^{(i)}} dk \, dl} = \frac{\displaystyle\int_{L^{(i)}} \left(\int_K |\mathrm{Disc}(l \cdot v_i)|^{1/6} |a(k l \cdot v_i)|^{-2/3} dk \right) dl} {\displaystyle\int_{L^{(i)}} dl} \label{yy} \end{equation} By Stanley Yao Xiao's answer, $\displaystyle\int_K |\mathrm{Disc}(l \cdot v_i)|^{1/6} |a(k l \cdot v_i)|^{-2/3} dk$ is equal to a constant $C$, independent of $l \cdot v_i$. Therefore, $(**)$ is equal to $C$.

Or, am I misconstruing?

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  • $\begingroup$ The follow-up question raised in the Addendum was answered here: mathoverflow.net/q/495129 $\endgroup$ Commented May 25 at 0:12

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We have the following theorem, dating back to work of Kurt Mahler in the 1930's:

Theorem: Let $f \in \mathbb{R}[x,y]$ be a binary form of degree $d \geq 3$ with non-zero discriminant $\Delta(f)$. Let

$$\displaystyle A_f = m(\{(x,y) \in \mathbb{R}^2 : |f(x,y)| \leq 1\}).$$

Here $m$ refers to Lebesgue measure on $\mathbb{R}^2$.

Then the quantity $A_f |\Delta(f)|^{\frac{1}{d(d-1)}}$ is a $\operatorname{GL}_2(\mathbb{R})$-invariant under the usual substitution action.

Now suppose $f$ is a real cubic form, with non-zero discriminant. We have

\begin{align*} A_f & = \iint_{\substack{(x,y) \in \mathbb{R}^2 \\ |f(x,y)| \leq 1}} dx dy \\ & = \iint_{\substack{(r, \theta) \in \mathbb{R}_{\geq 0} \times [0, 2\pi) \\ |f(r\cos \theta, r \sin \theta)| \leq 1}} r dr d\theta \\ & = \int_0^{2\pi} \int_0^{|f(\cos \theta, \sin \theta)|^{-1/3}} r dr d\theta \\ & = \int_{0}^{2 \pi} \frac{1}{2 |f(\cos \theta, \sin \theta)|^{2/3}} d \theta. \end{align*} Up to a constant, this is exactly what appears in the Bhargava-Shankar-Tsimerman paper.

This shows that

$$\displaystyle \frac{|\Delta(f)|^{1/6}}{2 \pi} \int_0^{2\pi} \frac{d\theta}{|f(\cos \theta, \sin \theta)|^{2/3}} $$

is $\operatorname{GL}_2(\mathbb{R})$-invariant, as claimed.

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    $\begingroup$ @user564434 you can look at the following paper by M.A. Bean: "An isoperimetric inequality for the area of plane regions defined by binary forms", Compositio Mathematica, Vol 92 (1994), 115-131 (numdam.org/article/CM_1994__92_2_115_0.pdf) $\endgroup$ Commented May 24 at 19:27
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    $\begingroup$ Excellent answer. Two small remarks. 1. Bhargava-Shankar-Tsimerman should have explained or referenced this. Moreover, they should have put an absolute value around $f(\cos \theta, \sin \theta)$ and earlier around $a(v)$ to disambiguate the $2/3$-power. 2. The proof of the claimed $\operatorname{GL}_2(\mathbb{R})$-invariance appears on p.119 of numdam.org/article/CM_1994__92_2_115_0.pdf $\endgroup$ Commented May 24 at 19:46
  • $\begingroup$ Thanks for your answer. I've added more to my original question in response. I would be grateful if you have any feedback. $\endgroup$ Commented May 24 at 21:58
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    $\begingroup$ @user564434 since the original question is resolved, it is better if you start a new question, especially since you have accepted my answer. For example, what if someone else addressed the second part of your edited question? Should you accept that answer instead? $\endgroup$ Commented May 24 at 22:03
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    $\begingroup$ @StanleyYaoXiao Okay. I have posted the follow-up as a separate question here $\endgroup$ Commented May 24 at 22:13

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