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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?

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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?

askedMay 24 at 10:33
user564434's user avatar
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  • $\begingroup$The follow-up question raised in the Addendum was answered here:mathoverflow.net/q/495129$\endgroup$CommentedMay 25 at 0:12

1 Answer1

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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.

answeredMay 24 at 19:04
Stanley Yao Xiao's user avatar
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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)$\endgroup$CommentedMay 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 ofnumdam.org/article/CM_1994__92_2_115_0.pdf$\endgroup$CommentedMay 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.$\endgroup$CommentedMay 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?$\endgroup$CommentedMay 24 at 22:03
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    $\begingroup$@StanleyYaoXiao Okay. I have posted the follow-up as a separate questionhere$\endgroup$CommentedMay 24 at 22:13

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