Gap Probabilities and Betti Numbers of a Random Intersection of Quadrics

Abstract

We consider the Betti numbers of an intersection of k random quadrics in $\mathbb {R}\text {P}^n$. Sampling the quadrics independently from the Kostlan ensemble, as $n \rightarrow \infty$ we show that for each $i\ge 0$ the expected ith Betti number satisfies

$$\begin{aligned} \mathbb {E}b_i(X)=1+O(n^{-M})\quad \text {for all } M>0. \end{aligned}$$

In other words, each fixed Betti number of X is asymptotically expected to be one; in fact as long as $i=i(n)$ is sufficiently bounded away from n / 2 the above rate of convergence is uniform (and in this range Betti numbers concentrate to their expected value). For the special case $k=2$ we study the expectation of the sum of all Betti numbers of X. It was recently shown (Lerario, in Proc. Am. Math. Soc. 143:3239–3251, 2015) that this expected sum equals $n+o(n)$; here we sharpen this asymptotic, showing that

$$\begin{aligned} \sum _{j=0}^n\mathbb {E}b_j(X)=n + \frac{2}{\sqrt{\pi }}n^{1/2} + O(n^c) \quad \text {for any } c>0 \end{aligned}$$

(1)

(the term $\tfrac{2}{\sqrt{\pi }}n^{1/2}$ comes from contributions of middle Betti numbers). The proofs are based on a combination of techniques from random matrix theory and spectral sequences. In particular (1) is based on a reduction that requires an average count of the number of singular quadrics in a random pencil; this count turns out to be related to the derivative at zero of the gap probability $f_{\beta , n}$ in finite Gaussian $\beta$-ensembles ($\beta =1,2,4$). We provide also new computations for this quantity and as n goes to infinity:

$$\begin{aligned} f_{\beta , n}'(0)\sim -\frac{2\sqrt{2}}{\pi }n^{1/2}. \end{aligned}$$

Appendix: Asymptotic Analysis of Theorem 11

The following lemma is a combination of Proposition 7 and Corollary 3 from [16] and gives the exact formula for $\mathcal {M}_{n-1}^{+}(1,2)$ together with its asymptotic behavior.

Lemma 14

$$\begin{aligned}&\mathcal {M}_{n-1}^{+}(1,2) \\&\quad = \frac{1}{2}\left\{ \begin{array}{cl} \frac{2\sqrt{2}}{\pi }\Gamma \left( \frac{n+1}{2}\right) &{} \text { for even } n,\\ (-1)^m\frac{(n-1)!}{m!2^{n-1}}+(-1)^{m-1} \frac{4\sqrt{2}(n-1)!}{\sqrt{\pi }m!2^{n-1}}\sum _{k=0}^{m-1}(-1)^k\frac{\Gamma (k+3/2)}{k!} &{} \text { for odd } n=2m+1. \\ \end{array} \right. \end{aligned}$$

Moreover as n goes to infinity (regardless its parity):

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(1,2)\sim \frac{\sqrt{2}}{\pi }\Gamma \big (\frac{n+1}{2}\big ). \end{aligned}$$

As for the other two cases we have the following lemmas.

Lemma 15

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3) = \left\{ \begin{array}{ll} \frac{1}{\pi }\Gamma \left( \frac{n+1}{2}\right) ^2 &{}\quad \text {for even } n,\\ \frac{n}{2\pi }\Gamma \left( \frac{n}{2}\right) ^2 &{}\quad \text {for odd } n=2m+1. \\ \end{array} \right. \end{aligned}$$

Moreover as n goes to infinity (regardless its parity):

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3)\sim \frac{1}{\pi }\Gamma \big (\frac{n+1}{2}\big )^2. \end{aligned}$$

Proof

We start by recalling the following formula from [26]:

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3)=\frac{1}{2}\prod _{j=1}^{n-1}\frac{\Gamma (3/2+\lfloor j/2\rfloor )}{\Gamma (1/2+\lfloor j/2\rfloor )}. \end{aligned}$$

Using the identity $\Gamma (z+1)=z\Gamma (z)$ in the above formula with $z=1/2+\lfloor j/2\rfloor ,$ we can rewrite it as

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3)&=\frac{1}{2}\prod _{j=1}^{n-1}\big (\frac{1}{2} + \big \lfloor \frac{j}{2}\big \rfloor \big )\\&=\frac{1}{2}\big (\prod _{1\le j\le n-1, j \text { even}}\frac{j+1}{2}\big )\cdot \big (\prod _{1\le j\le n-1, j \text { odd}}\frac{j}{2}\big )\\&=\frac{1}{2^{n}}\big (\prod _{2\le k\le n, k \text { odd}}k\big )\cdot \big (\prod _{1\le k\le n, k \text { odd}}k\big )=\frac{c_n}{2^n}\prod _{1\le k\le n, k \text { odd}}k^2, \end{aligned}$$

where $c_n=1$ for even n and n for odd ones. Thus if $n=2m$ is even, we have

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3)&=\frac{1}{2^n}(2m-1)!!^2\\&=\frac{2^{2m-n}}{\pi }\Gamma (m+1/2)^2 = \frac{1}{\pi } \Gamma \big (\frac{n+1}{2}\big )^2, \end{aligned}$$

where in the last line we have used the identity $\Gamma (m+1/2)=\sqrt{\pi }\frac{(2m-1)!!}{2^m}.$ In the case $n=2m+1$ is odd, recalling the value $c_{n\text { odd}}=n$, we have

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(2,3)&=n\frac{1}{2^n}(1\cdot 2\cdots (2m-1))^2 = n \frac{\Gamma (m+1/2)^2}{2\pi }\\&=\frac{n}{2\pi }\Gamma \left( \frac{n}{2}\right) ^2. \end{aligned}$$

The asymptotics are a simple application of Stirling’s formula. $\square$

Lemma 16

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(4,5)=\frac{4^{-n+1}}{\pi } \Gamma \big ( n+ \frac{1}{2} \big )^2 2 H_n(-1) , \end{aligned}$$

where $H_n$ is a hypergeometric function such that $2 H_n(-1) = 1 + o(1)$ as $n \rightarrow \infty$. In particular as n goes to infinity:

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(4,5)\sim \frac{1}{4^{n-1}\pi }\Gamma \big (n+\frac{1}{2}\big )^2. \end{aligned}$$

Proof

We start by recalling equation (26.3.10) from [26]:

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(4,5)&=\frac{1}{2^{2n-1}}\prod _{j=0}^{n-2} \frac{\Gamma (j+5/2)}{\Gamma (j+3/2)}\sum _{k=0}^{n-1}{n-1\atopwithdelims ()k}(1)_{k}(3/2)_{n-1-k}\\&=\frac{1}{2^{2n-1}}\big (\prod _{j=0}^{n-2} \frac{\Gamma (j+5/2)}{\Gamma (j+3/2)}\big )\cdot \frac{2}{\pi }\Gamma (n+1/2) {}_{2}F_{1} \big ( 1,1-n,\frac{1}{2}-n,-1 \big ). \end{aligned}$$

In the above line $_2F_1$ denotes the hypergeometric function; let us set

$$\begin{aligned} H_{n}(-1)= {}_{2}F_{1} \big ( 1,1-n,\frac{1}{2}-n,-1 \big ). \end{aligned}$$

With this notation we have

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(4,5)&=\frac{2 H_{n}(-1)}{\sqrt{\pi }2^{2n-1}} \Gamma (n+1/2)\prod _{j=0}^{n-2} \frac{\Gamma (j+5/2)}{\Gamma (j+3/2)}\\&=\frac{2 H_{n}(-1)}{\sqrt{\pi }2^{2n-1}} \Gamma (n+1/2)\prod _{j=0}^{n-2}(j+3/2), \end{aligned}$$

where again the last line we have used $\Gamma (z+1)=z\Gamma (z)$ for $z=j+3/2.$ Recalling also the identity $\prod _{j=0}^{n-2}(j+3/2)=\frac{2}{\pi }\Gamma (n+1/2),$ we finally get

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(4,5)=\frac{2 H_{n}(-1)}{\pi 4^{n-1}}\Gamma \big (n+\frac{1}{2}\big )^2. \end{aligned}$$

It remains to prove the limit $2H_n(-1)\rightarrow 1$. First we use the Pfaff transformation:

$$\begin{aligned} {}_{2}F_{1}(a,b;c;z) = (1-z)^{-a} \cdot {}_{2}F_{1} \big ( a,b-c;c;\frac{z}{z-1} \big ). \end{aligned}$$

In our case

$$\begin{aligned} 2 \cdot {}_{2}F_{1} \big ( 1,-n;-n-\frac{1}{2};-1 \big ) = {}_{2}F_{1} \big ( 1,-\frac{1}{2};-n-\frac{1}{2};\frac{1}{2} \big ). \end{aligned}$$

Now we use the series definition in terms of the Pockhammer symbol:

$$\begin{aligned} {}_{2}F_{1}\big ( 1,-\frac{1}{2};-n-\frac{1}{2};\frac{1}{2} \big ) = \sum _{k=0}^{\infty } \frac{(-1/2)_k}{(-n-1/2)_k} (1/2)^k. \end{aligned}$$

We need to show that the right hand side $\rightarrow 1$. We have

$$\begin{aligned} \sum _{k=0}^{\infty } \frac{(-1/2)_k}{(-n-1/2)_k} (1/2)^k&= 1 + \frac{1}{4(n+1/2)}\sum _{k=1}^{\infty } \frac{(1/2)_{k-1}}{(-n+1/2)_{k-1}} (1/2)^{k-1}. \end{aligned}$$

(25)

We use the rough bound:

$$\begin{aligned} \big | \sum _{k=0}^{\infty } \frac{(1/2)_{k}}{(-n+1/2)_{k}} (1/2)^{k} \big |&\le \sum _{k=0}^{n} (1/2)^{k} + \frac{1}{|(-n+1/2)_n|} \sum _{k=n+1}^{\infty } \frac{(1/2)_{k}}{(1/2)_{k-n}} (1/2)^{k}, \\&\le 2 + \frac{(1/2)^n}{|(-n+1/2)_n|} \sum _{k=n+1}^{\infty } \frac{k!}{(k-n)!} (1/2)^{k-n} \\&= 2+ \frac{n!}{|(-n+1/2)_n|}(1/2)^n \frac{F^{(n)}(1/2)}{n!} , \end{aligned}$$

where

$$\begin{aligned} F(z) = \frac{1}{1-z} . \end{aligned}$$

We have

$$\begin{aligned} F^{(n)}(1/2)/n! = O(1) \cdot 2^n . \end{aligned}$$

Applying this along with Stirling’s approximation:

$$\begin{aligned} 2+ \frac{n!}{|(-n+1/2)_n|}(1/2)^n \frac{F^{(n)}(1/2)}{n!} = 2 + \frac{n!}{|(-n+1/2)_n|} O(1) = o(n). \end{aligned}$$

This shows that (25) equals $1 + \frac{1}{4(n+1/2)} o(n) = 1 + o(1)$, as desired. $\square$

As a corollary we get the following asymptotic for the volume of $\Sigma _{\beta ,n}$.

Corollary 17

For each $\beta =1,2,4$ we have

$$\begin{aligned} \frac{|\Sigma _{\beta ,n} |}{|S^{N_\beta - 2}|}\sim \frac{2}{\sqrt{\pi }}n^{\frac{1}{2}}. \end{aligned}$$

Remark 4

In our main application of this asymptotic (Theorem 8) we will need, for $\beta =1$, a more precise error bound:

$$\begin{aligned} \frac{|\Sigma _{1,n} |}{|S^{N_1 - 2}|} = \frac{2}{\sqrt{\pi }}n^{\frac{1}{2}} + O(1). \end{aligned}$$

Proof

Recall from Theorem 11 (and the remark below it) that

$$\begin{aligned} \frac{|\Sigma _{\beta ,n} |}{|S^{N_\beta - 2}|} =2n \beta ^{\frac{n\beta -\beta +1}{2}}\frac{\Gamma (1+\beta /2)}{\Gamma (1+\beta n/2)}\mathcal {M}_{n-1}^{+}(\beta , \beta +1). \end{aligned}$$

The result follows applying Stirling’s approximation to the asymptotic for $\mathcal {M}_{n-1}^{+}(\beta , \beta +1)$ given in Lemmas 14, 15, and 16.

The error bound stated in the Remark follows immediately from the error in Stirling’s approximation for n even. For $n = 2m+1$ odd, reading the proof of Lemma 14 which was given in [16, Cor. 3], one can conclude that

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(1,2) = \frac{(n-1)!}{m! 2^{n-1}} \big ( (-1)^m + \frac{4\sqrt{2}}{\sqrt{\pi }} S_m \big ), \end{aligned}$$

(26)

where

$$\begin{aligned} S_m = \frac{1}{2}\sum _{j=0}^{m/2-1} \frac{\Gamma (2j + 2 - 1/2)}{ \Gamma (2j + 2)}, \end{aligned}$$

for m even, and

$$\begin{aligned} S_m = \frac{\Gamma (m+1/2)}{\Gamma (m)} - \frac{1}{2}\sum _{j=0}^{(m-1)/2-1} \frac{\Gamma (2j + 2 - 1/2)}{ \Gamma (2j + 2)}, \end{aligned}$$

for m odd. Using the asymptotic [29]

$$\begin{aligned} \frac{\Gamma (z+a)}{\Gamma (z+b)} = z^{a-b}(1+ O(1/z)), \end{aligned}$$

along with an integral estimate for the sum we have (regardless of the parity of m):

$$\begin{aligned} S_m = \frac{\sqrt{m}}{2} + O(m^{-1/2}). \end{aligned}$$

Applying this to (26) gives

$$\begin{aligned} \mathcal {M}_{n-1}^{+}(1,2) = \frac{(n-1)!}{m! 2^{n-1}} \big ( (-1)^m + \frac{4\sqrt{2}}{\sqrt{\pi }} S_m \big ) = \frac{\sqrt{2}}{\pi } \Gamma \big ( \frac{n+1}{2} \big ) ( 1 + O(n^{-1/2})). \end{aligned}$$

Using Stirling’s approximation for

$$\begin{aligned} \frac{C_1(n)}{C_1(n-1)} = \frac{\Gamma (1+1/2)}{\sqrt{2\pi } \Gamma (1+n/2)}, \end{aligned}$$

and plugging this into the exact formula gives:

$$\begin{aligned} \frac{|\Sigma _{1,n} |}{|S^{N_1 - 2}|} = \frac{2}{\sqrt{\pi }}n^{\frac{1}{2}}(1 + O(n^{-1/2})). \end{aligned}$$

The asymptotic of Theorem 10 follows again from Lemmas 14, 15 and 16.

Corollary 18

The following asymptotic holds for the derivative at zero of the gap probability:

$$\begin{aligned} f'_{\beta , n}(0)\sim -\frac{2\sqrt{2}}{\pi }n^{1/2}. \end{aligned}$$

$\square$