TITLE: Condition number for a symmetric positive definite matrix
QUESTION [6 upvotes]: I am trying to get an estimate for the induced 2-norm condition number $\kappa_{2}(M)$ of this matrix $M$:
$$M_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x,$$
which is symmetric positive definite of size $n\times n$. The above link gives the elements of the inverse. We also know that its determinant is the reciprocal of this OEIS sequence.
I want to obtain something like $\kappa_{2}(M) = \Omega(f(n))$ for some to-be-determined $f(n)$.
From numerics, it seems that $\kappa_2$ grows very fast with respect to $n$. But I am not sure what might be an analytic way to obtain the asymptotic for $\lambda_{\max}/\lambda_{\min}$.

REPLY [1 votes]: (Too long for a comment.)
I managed to numerically extract the condition numbers up to $n = 112$. I have plotted this on a log scale:

It looks a bit faster than linear on a log scale; maybe a small quadratic term? But it looks no faster than $e^{n^2}$. The best fit line is dashed. The best fit parabola of the natural logarithm of the condition number is $-33.5838 + 8.09351n + 0.0217842n^2$. I also tried fitting a cubic polynomial as a sanity check, obtaining $-19.6223 + 6.71268n + 0.0516761n^2 - 0.000174806n^3$; that the coefficient of $n^3$ is negative and small might be indicative that it isn't "real", but of course we can't rule that out numerically.
Code:
#include <Eigen/Dense>
#include <Eigen/Eigenvalues>
#include <boost/math/special_functions/factorials.hpp>
#include <boost/multiprecision/cpp_bin_float.hpp>
#include <iostream>
#include <vector>
#include <cmath>
#include <fstream>

using Real = boost::multiprecision::number<boost::multiprecision::cpp_bin_float<1000>>;
using Mat = Eigen::Matrix<Real, Eigen::Dynamic, Eigen::Dynamic>;
using boost::math::factorial;
using std::log;
using std::sqrt;

int main() {
    std::ofstream of{"data.csv"};
    std::cout << std::setprecision(std::numeric_limits<Real>::digits10);
    of << "n, cond(M)\n";
    for (int n = 2; n < 200; ++n) {
        std::cout << "n = " << n << "\n";
        Mat X = Mat::Zero(n,n);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                Real d3 = Real(2*n - i - j - 1);
                X(i,j) = 1/(factorial<Real>(n - i - 1)*factorial<Real>(n - j - 1)*d3);
            }
        }
        Eigen::SelfAdjointEigenSolver<Mat> es(n);
        es.compute(X);

        std::vector<Real> v(es.eigenvalues().data(), es.eigenvalues().data() + n);
        if (!std::is_sorted(v.begin(), v.end())) {
            std::cerr << "Expected invariant is broken.\n";
            return 1;
        }
        assert(v.front() > 0);

        Real cond = v.back()/v.front();
        if (cond > 0.01/sqrt(std::numeric_limits<Real>::epsilon())) {
            std::cerr << "Precision must be increased to get more samples; fails at n = " << n << "\n";
            break;
        }
        of << n << ", " << cond << std::endl;
    }
    of.close();
}

Generated data:
n, cond(M)
2, 19.2815
3, 1181.56
4, 165823
5, 4.18166e+07
6, 1.65669e+10
7, 9.47936e+12
8, 7.39574e+15
9, 7.54511e+18
10, 9.7498e+21
11, 1.55626e+25
12, 3.00702e+28
13, 6.91676e+31
14, 1.86767e+35
15, 5.84992e+38
16, 2.10375e+42
17, 8.60899e+45
18, 3.97753e+49
19, 2.06044e+53
20, 1.18933e+57
21, 7.60721e+60
22, 5.36477e+64
23, 4.15244e+68
24, 3.51305e+72
25, 3.2363e+76
26, 3.23507e+80
27, 3.49783e+84
28, 4.07854e+88
29, 5.11458e+92
30, 6.88024e+96
31, 9.90493e+100
32, 1.52261e+105
33, 2.49406e+109
34, 4.34471e+113
35, 8.03433e+117
36, 1.57442e+122
37, 3.26413e+126
38, 7.14851e+130
39, 1.65132e+135
40, 4.01803e+139
41, 1.02846e+144
42, 2.76577e+148
43, 7.80505e+152
44, 2.30875e+157
45, 7.15071e+161
46, 2.31656e+166
47, 7.84207e+170
48, 2.7714e+175
49, 1.02155e+180
50, 3.92398e+184
51, 1.56944e+189
52, 6.5308e+193
53, 2.82524e+198
54, 1.26967e+203
55, 5.92328e+207
56, 2.86664e+212
57, 1.43826e+217
58, 7.4761e+221
59, 4.02365e+226
60, 2.24086e+231
61, 1.29065e+236
62, 7.68353e+240
63, 4.72539e+245
64, 3.00063e+250
65, 1.96638e+255
66, 1.3292e+260
67, 9.26348e+264
68, 6.65305e+269
69, 4.92194e+274
70, 3.74915e+279
71, 2.93919e+284
72, 2.37052e+289
73, 1.96612e+294
74, 1.67632e+299
75, 1.46866e+304
76, 1.32174e+309
77, 1.22145e+314
78, 1.15867e+319
79, 1.12785e+324
80, 1.12618e+329
81, 1.15317e+334
82, 1.21051e+339
83, 1.30226e+344
84, 1.43536e+349
85, 1.62041e+354
86, 1.87313e+359
87, 2.21652e+364
88, 2.68421e+369
89, 3.32574e+374
90, 4.21478e+379
91, 5.46219e+384
92, 7.23698e+389
93, 9.80034e+394
94, 1.35617e+400
95, 1.91725e+405
96, 2.76844e+410
97, 4.08212e+415
98, 6.14524e+420
99, 9.44279e+425
100, 1.48074e+431
101, 2.36913e+436
102, 3.86672e+441
103, 6.43655e+446
104, 1.09254e+452
105, 1.89067e+457
106, 3.33506e+462
107, 5.99552e+467
108, 1.09826e+473
109, 2.04958e+478
110, 3.89608e+483
111, 7.54267e+488
112, 1.4869e+494

(These number break every plotting program I've tried past $n = 75$ as that's where they start overflowing double.)


TITLE: Solving for $g(y)$ in the following integral equation
QUESTION [1 upvotes]: Given the following integral equation where $f, h$ are known:
$$ f(t) = \int_{t-1}^t{g(y)\cdot h(y)\ \textrm{d}y}$$
Is it possible to solve this for $g(t)$, i.e. get an equation of the form $g(t) = \dots$ (supposing $g, h$ are suitably nice)? I have no idea where to start. I acknowledge that an analytical solution may not even exist in the general case, but I remain hopeful!

REPLY [1 votes]: ok lets suppose we have some function: $z(y)=g(y)h(y)$ then we can say:
$$f(t)=\int_{t-1}^tz(y)dy$$
now suppose that $z$ satisfies the equation $Z'(y)=z(y)$ so we have:
$$f(t)=Z(t)-Z(t-1)$$
now if we differentiate both sides we get:
$$f'(t)=Z'(t)-Z'(t-1)=z(t)-z(t-1)$$
now substituting back in we get:
$$f'(t)=g(t)h(t)-g(t-1)h(t-1)$$
how if we isolate out $g(t)$ we get:
$$g(t)=\frac{f'(t)+g(t-1)h(t-1)}{h(t)}$$
and so:
$$g(t)=\frac{h(t-1)}{h(t)}\color{red}{g(t-1)}+\frac{f'(t)}{h(t)}$$
so whilst we cannot get a pure solution for $g(t)$ we can get this recurrance relation style which I believe you could get a solution from using generating functions depending on what the other functions are. Hope this helps


TITLE: Calculate Magnetic Flux Density between Two Disc Magnets
QUESTION [0 upvotes]: If I have two disc-shaped magnets (radius r=0.05m, width w=0.03m, Remanance Br=1.06 T) separated by a distance d, how can I calculate the magnetic flux density somewhere between them?
I found this webpage but I get a tiny value for B (10^-4 T) and I also have no idea where they get the formula from and whether it is trust-worthy.

REPLY [0 votes]: The given formula (in that website) is correct.
Assume we have a cylinder with permanent (axial) magnetization $\mathbf{M}$. It will cause a surface current density $\mathbf{K}$:
$$\cases{\bf{K}=\bf M \times \hat n \\ \mathbf M=M_0\hat z}\to \mathbf{K}=M_0\hat \phi$$
So it is like a finite length solenoid. To find the field on it's axis ($z$, point $P$ in the below picture), from Biot–Savart law we'll arrive at:

$$dB=dB_z=\frac{\mu_0K\mathrm{d}z}{2}\frac{R^2}{\xi^2}$$
$$\cases{dz=\frac{Rd\theta}{\sin^2\theta}\\ \frac{R}{z}=\tan \theta \\ \xi=\frac{R}{\sin \theta}}\to dB=-\frac{\mu_0 K}{2}\sin \theta d\theta$$
$$B=\int_{\theta_1}^{\theta_2}dB=\frac{\mu_0 K}{2}(\cos \theta_1-\cos \theta_2)$$
 so
$$\boxed {\mathbf{B}=\frac{\mu_0 M_0}{2}(\cos \theta_1-\cos \theta_2)\hat z}$$


text:
\begin{document}
\renewcommand{\abstractname}{Abstract}
\begin{abstract}
In this note we show the following result using the integral-geometric formula of R. Howard: Consider the totally geodesic $\mathbb{R}P^{2m}$ in $\mathbb{C}P^n$. Then it minimizes volume among the isotropic submanifolds in the same $\mathbb{Z}/2$ homology class in $\mathbb{C}P^n$ (but not among all submanifolds in this $\mathbb{Z}/2$ homology class). Also the totally geodesic $\mathbb{R}P^{2m-1}$ minimizes volume in its Hamiltonian deformation class in $\mathbb{C}P^n$. As a corollary we'll give estimates for volumes of Lagrangian submanifolds in complete intersections in $\mathbb{C}P^n$.
\end{abstract}
\title[Isotropic submanifolds...]{Volume minimization and estimates for certain isotropic submanifolds in complex projective spaces}
\author{Edward Goldstein}
\maketitle

\section{Introduction}
On a K\"ahler $n$-fold $M$ there is a class of {\it isotropic} submanifolds. Those are submanifolds of $M$ on which the K\"ahler form $\omega$ of $M$ vanishes. The maximal dimension of such a submanifold is $n$ (the middle dimension) in which case it is called {\it Lagrangian}.\\
In this papers we'll exhibit global volume-minimizing properties among isotropic competitors for certain submanifolds of the complex projective space. In general global volume-minimizing properties of minimal/Hamiltonian stationary Lagrangian/isotropic submanifolds in K\"ahler (particularly K\"ahler-Einstein) manifolds are still poorly understood. In dimesion $2$ there is a result of Schoen-Wolfson \cite{ScW} (extended to isotropic case by Qiu in \cite{Qiu}) which shows existence of Lagrangian cycles minimizing area among Lagrangians in a given homology class. Still it is not clear whether a {\it given} minimal Lagrangian has any global volume-minimizing properties.\\
The only instance where we have a clear cut answer to global volume-minimizing problem is Special Lagrangian submanifolds which are homologically volume-mimizing in Calabi-Yau manifolds \cite{HaL}. In K\"ahler-Einstein manifolds of negative scalar curvature, besides geodesics on Riemann surfaces of negative curvature, we have some examples \cite{Lee} of minimal Lagrangian submanifolds which are homotopically volume-minimizing. The author has a program for studying homotopy volume-minimizing properties for Lagrangians in K\"ahler-Einstein manifolds of negative scalar curvature \cite{Gold1}, but so far there are no satisfactory results.\\
In positive curvature case there is a result of Givental-Kleiner-Oh which states that the canonical totally geodesic $\mathbb{R}P^{n}$ in $\mathbb{C}P^{n}$ minimizes volume in its Hamiltonian deformation class, \cite{Giv}. The proof uses integral geometry and Floer homology to study intersections for Hamiltonian deformations of $\mathbb{R}P^{n}$. Those arguments can be generalized to products of Lagrangians in a product of symmetric K\"ahler manifolds, \cite{IOS}. There is a related conjecture due to Oh that the Clifford torus minimizes volume in its Hamiltonian deformation class in $\mathbb{C}P^{n}$, \cite{Oh}. Some progress towards this was obtained in \cite{Gold2}. Also general lower bounds for volumes of Lagrangians in a given Hamiltonian deformation class in $\mathbb{C}^n$ were obtained in \cite{Vit}.\\
In this note we extend and improve the result of Givental-Kleiner-Oh to isotropic totally geodesic $\mathbb{R}P^k$ sitting canonically in $\mathbb{C}P^n$. Our main result is the following theorem:
\begin{tom}
Consider the totally geodesic $\mathbb{R}P^{2m}$ in $\mathbb{C}P^n$. Then it minimizes volume among the isotropic submanifolds in the same $\mathbb{Z}/2$ homology class in $\mathbb{C}P^n$ (but not among all submanifolds in this $\mathbb{Z}/2$ homology class). Also consider the totally geodesic $\mathbb{R}P^{2m-1}$ in $\mathbb{C}P^n$. Then it minimizes volume in its Hamiltonian deformation class.
\end{tom}
A corollary of this is:
\begin{corol}
Let $f_1,\ldots,f_k$ be real homogeneous polynomials of odd degree in $n+1$ variables with $2m+k=n$. Let $N$ be the zero locus of $f_i$ in $\mathbb{C}P^n$ and $L$ be their real locus. Then $vol(L) \leq \Pi deg(f_i) vol(\mathbb{R}P^{2m})$ and if $L'$ is a Lagrangian submanifold of $N$ homologous mod $2$ to $L$ in $N$ then $vol(L') \geq vol(\mathbb{R}P^{2m})$.
\end{corol}
\section{A formula from integral geometry}
In this section we establish a formula from integral geometry for volumes of isotropic submanifolds of $\mathbb{C}P^n$ following the exposition in R. Howard \cite{How}.\\
In our case the group $SU(n+1)$ acts on $\mathbb{C}P^n$ with a stabilizer $K \simeq U(n)$. Thus we view $\mathbb{C}P^{n}=SU(n+1)/K$ and the Fubini-Study metric is induced from the bi-invariant metric on $SU(n+1)$. Let $P^{2m}$ be an isotropic submanifold of $\mathbb{C}P^n$ of dimension $2m$ and let $Q$ be a linear $\mathbb{C}P^{n-m} \subset \mathbb{C}P^n$. For a point $p \in P$ and $q \in Q$ we define an angle $\sigma(p,q)$ between the tangent planes $T_pP$ and $T_qQ$ as follows: First we choose some elements $g$ and $h$ in $SU(n+1)$ which move $p$ and $q$ respectively to the same point $r \in \mathbb{C}P^n$. Now the tangent planes $g_{\ast}T_pP$ and $h_{\ast}T_qQ$ are in the same tangent space $T_r \mathbb{C}P^n$ and we can define an angle between them as follows: take an orthonormal basis $u_1\ldots u_{2m}$ for $g_{\ast}T_pP$ and an orthonormal basis $v_1 \ldots v_{2n-2m}$ for $h_{\ast}T_qQ$ and define \[\sigma(g_{\ast}T_pP,h_{\ast}T_qQ)= |u_1 \wedge \ldots \wedge v_{2n-2m}|\]
The later quantity $\sigma(g_{\ast}T_pP,h_{\ast}T_qQ)$ depends on the choices $g$ and $h$ we made. To mend this we'll need to average this out by the stabilizer group $K$ of the point $r$. Thus we define:
\[\sigma(p,q)=\int_{K} \sigma(g_{\ast}T_pP, k_{\ast}h_{\ast}T_qQ) dk\]
Since $SU(n+1)$ acts transitively on the Grassmanian of isotropic planes and the complex planes in $\mathbb{C}P^n$ we conclude that this angle is a constant depending just on $m$ and $n$:
\[\sigma(p,q)=C_{m,n}\]
There is a following general formula due to R. Howard \cite{How}:
\[ \int_{SU(n+1)} \#(P \bigcap gQ) dg= \int_{P \times Q} \sigma(p,q) dp dq= C_{m,n} vol(P)vol(Q)\]
Here $\#(P \bigcap gQ)$ is the number of intersection points of $P$ with $gQ$, which is finite for a generic $g \in SU(n+1)$. To use the formula we need to have some control over the intersection pattern of $P$ and $gQ$. We have the following lemma:
\begin{lem}
\label{tech}
Let $P$ be the totally geodesic $\mathbb{R}P^{2m} \subset \mathbb{C}P^n$, let $Q=  \mathbb{C}P^{n-m} \subset \mathbb{C}P^n$. Let $g \in SU(n+1)$ s.t. $P$ and $gQ$ intersect transversally. Then $ \#(P \bigcap gQ)=1$. Also let $f_1,\ldots,f_k$ be real homogeneous polynomials in $n+1$ variables with $2m+k=n$ and let $P'$ be their real locus. If $P'$ is transversal to $gQ$ then $ \#(P' \bigcap gQ) \leq \Pi deg(f_i)$.
\end{lem}
{\bf Proof:} For the first claim we have $gQ$ is given by an $(n-m+1)$-plane $H \subset \mathbb{C}^{n+1}$ and hence it is a zero locus of $m$ linear equations on $\mathbb{C}^{n+1}$. Hence $(P \bigcap gQ)$ is cut out by $2m$ linear equations in $\mathbb{R}P^{2m}$.\\
For the second claim we note that as before $gQ \bigcap \mathbb{R}P^n$ is the zero locus of $2m$ linear polymonials $h_1,\ldots, h_{2m}$ on $\mathbb{R}P^n$. Moreover $P'$ is a zero locus of $f_1,\ldots,f_{n-2m}$ on $\mathbb{R}P^n$. For generic $g \in SU(n+1)$ we'll have that $gQ$ and $P'$ intersect transversally in $\mathbb{R}P^n$. By Bezout's theorem (see \cite{GH}, p. 670) the common zero locus of $h_1,\ldots, h_{2m}$ and $f_1,\ldots,f_{n-2m}$ is $\mathbb{C}P^n$ is $\Pi deg(f_i)$ points. Now $P' \bigcap gQ$ is a part of this locus, hence $ \#(P' \bigcap gQ) \leq \Pi deg(f_i)$.

\section{Proof of the volume minimization}

Now we can prove the result stated in the Introduction:
\begin{thm}
Consider the totally geodesic $\mathbb{R}P^{2m}$ in $\mathbb{C}P^n$. Then it minimizes volume among the isotropic submanifolds in the same $\mathbb{Z}/2$ homology class in $\mathbb{C}P^n$ (but not among all submanifolds in this $\mathbb{Z}/2$ homology class). Also consider the totally geodesic $\mathbb{R}P^{2m-1}$ in $\mathbb{C}P^n$. Then it minimizes volume in its Hamiltonian deformation class.
\end{thm}
{\bf Proof:} Let $P$ be an isotropic submanifold homologous to $\mathbb{R}P^{2m}$ mod $2$ and let $Q=\mathbb{C}P^{n-m}$. By Lemma \ref{tech} the intersection number mod $2$ of $P$ and $gQ$ is $1$. Hence the formula in the previous section tells that
\[C_{m,n} vol(P)vol(Q)= \int_{SU(n+1)} \#(P \bigcap gQ) dg \geq vol(SU(n+1))\]
and
\[C_{m,n} vol(\mathbb{R}P^{2m})vol(Q)= \int_{SU(n+1)} \#(\mathbb{R}P^{2m} \bigcap gQ) dg = vol(SU(n+1))\]
and this proves the first part. We also note that that $\mathbb{C}P^1$ is homologous to $\mathbb{R}P^2$ mod $2$ in $\mathbb{C}P^n$ but \[vol(\mathbb{C}P^1) <vol(\mathbb{R}P^2) \]
The second assertion will follow from the first one. Consider $\mathbb{C}^{n+1}$ and a unit sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$. We have a natural circle action on $S^{2n+1}$ (multiplication by unit complex numbers). Let the vector field $u$ be the generator of this action. We have a $1$-form $\alpha$ on $S^{2n+1}$, \[ \alpha(v) = u \cdot v \] Also $d\alpha=2\omega$ where $\omega$ is the K\"ahler form of $\mathbb{C}^{n+1}$. The kernel of $\alpha$ is the {\it horizontal distribution}. We have a Hopf map $\rho: S^{2n+1} \mapsto \mathbb{C}P^n$. We have $\mathbb{R}P^{2m-1} \subset \mathbb{C}P^n$ and $S^{2m-1} \subset  S^{2n+1}$ which is a horizontal double cover of  $\mathbb{R}P^{2m-1}$. \\
Let $f$ be a (time-dependent) Hamiltonian function on  $\mathbb{C}P^n$. Then we can lift it to a Hamiltonian function on $\mathbb{C}^{n+1}-(0)$ and its Hamiltonian vector field $H_f$ is horizontal on $S^{2n+1}$. Consider now the vector field \[w=-2f \cdot u+H_f\]
The vector field $w$ is $S^1$-invariant. We also have:
\begin{prop}
The Lie derivative $L_{w}\alpha= 0$
\end{prop}
{\bf Proof:} We have \[L_{w}\alpha= d(i_w \alpha)+i_w d \alpha= -2df+2df\]
Let now $\Phi_t$ be the time $t$ flow of $w$ on $S^{2n+1}$ and let $\Xi_t$ be the Hamiltonian flow of $f$ on $\mathbb{C}P^n$. Then $\Phi_t(S^{2m-1})$ is horizontal and isotropic and it is a double cover of $\Xi_t(\mathbb{R}P^{2m-1})$. Hence
\[vol(\Phi_t(S^{2m-1}))=2vol(\Xi_t(\mathbb{R}P^{2m-1})) \]
Let $S_t=\Phi_t(S^{2m-1})$. We build a suspension $\Sigma S_t$ of $S_t$ in $S^{2n+3} \subset \mathbb{C}^{n+2}$, \[\Sigma S_t = \big((\sin \theta \cdot x, \cos \theta) \in  \mathbb{C}^{n+2}=  \mathbb{C}^{n+1} \oplus  \mathbb{C}| 0 \leq \theta \leq \pi ~ , ~  x \in S_t \big)\]
One immediately verifies that $\Sigma S_t$ is horizontal and it is a double cover of an isotropic submanifold $L_t$ (with a conical singularity) of $\mathbb{C}P^{n+1}$ with $L_0=\mathbb{R}P^{2m}$. Also one readily checks that \[vol(\Sigma S_t)= vol(S_t) \cdot \int_{\theta=0}^{\pi} \sin^{2m-1} \theta \ d \theta\]
Hence
\[2 vol(L_t)=vol(\Sigma S_t)=  2vol(\Xi_t(\mathbb{R}P^{2m-1})) \cdot \int_{\theta=0}^{\pi} \sin^{2m-1} \theta \ d \theta \]
Now the first part of our theorem implies that $vol(L_t) \geq vol(L_0)$. Hence we conclude that $vol(\Xi_t(\mathbb{R}P^{2m-1})) \geq vol(\mathbb{R}P^{2m-1})$.
 Q.E.D. \\
{\bf Remark:} One notes from the proof that for $\mathbb{R}P^{2m-1}$ it would be suffient to use exact deformations by isotropic immersions of
$\mathbb{R}P^{2m-1}$. A family $L_t$ of isotropic immersions of
$\mathbb{R}P^{2m-1}$ is called {\it exact} if the $1$-form $i_v \omega$ is exact when restricted to each element of the family. Here $v$ is the deformation vector field and $\omega$ is the symplectic form. Thus embeddedness is not important for the conclusion of the theorem.\\
The theorem has the following corollary:
\begin{cor}
Let $f_1,\ldots,f_k$ be real homogeneous polynomials of odd degree in $n+1$ variables with $2m+k=n$. Let $N$ be the zero locus of $f_i$ in $\mathbb{C}P^n$ and $L$ be their real locus. Then $vol(L) \leq \Pi deg(f_i) vol(\mathbb{R}P^{2m})$ and if $L'$ is a Lagrangian submanifold of $N$ homologous mod $2$ to $L$ in $N$ then $vol(L') \geq vol(\mathbb{R}P^{2m})$.
\end{cor}
{\bf Proof:} We note that $N$ is a complex $2m$-fold and $L$ is its Lagrangian submanifold. Since the degrees of $f_i$ are odd, we have by adjunction formula that $L$ and $\mathbb{R}P^{2m}$ represent the same homology class in $H_{2m}(\mathbb{R}P^n,\mathbb{Z}/2)$. Let $Q$ be a linear $\mathbb{C}P^{n-m}$ in $\mathbb{C}P^n$ and $g \in SU(n+1)$. The intersection munber mod $2$ of $gQ$ with $L'$ is $1$. We have that
\[C_{m,n} vol(\mathbb{R}P^{2m})vol(Q)= \int_{SU(n+1)} 1 dg\]
\[C_{m,n} vol(L')vol(Q)= \int_{SU(n+1)} \#(L' \bigcap gQ) dg \]
Also using Lemma \ref{tech}:
\[C_{m,n} vol(L)vol(Q)= \int_{SU(n+1)} \#(L \bigcap gQ) dg \leq \Pi deg(f_i) vol(SU(n+1))\] and our claims follow.  Q.E.D. \\

config: arxiv