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Projective transformations of convex bodies and volume product

F. Balacheff, G. Solanes & K. Tzanev
Abstract.

In this paper we study certain variational aspects of the volume product functional restricted to the space of small projective deformations of a fixed convex body. In doing so, we provide a short proof of a theorem by Klartag: a strong version of the slicing conjecture implies the Mahler conjecture. We also exhibit an interesting family of critical convex bodies in dimension 22 containing saddle points for this functional.

Key words and phrases:
Convex body, Mahler conjecture, slicing conjecture, volume product.
2010 Mathematics Subject Classification:
Primary: 52A20, 52A40. Secondary: 52A38
The first author acknowledges support by the FSE/AEI/MICINN grant RYC-2016-19334 “Local and global systolic geometry and topology”. The second author is supported by the Serra Hunter Programme and the AEI/MICINN María de Maeztu grant CEX2020-001084-M. The first and the second authors acknowledge support by the FEDER/AEI/MICINN grant PID2021-125625NB-I00 “Estructuras y Desigualdades Geométricas Universales” and the AGAUR grant 2021-SGR-01015.

1. Introduction

The volume product of a convex body KnK\subset\mathbb{R}^{n} whose interior contains the origin is the quantity vp(K)|K||K|\operatorname{vp}(K)\coloneqq|K|\cdot|K^{\circ}|. Here |||\cdot| denotes the standard Lebesgue measure in n\mathbb{R}^{n}, and K={ynx,y=i=1nxiyi1xK}K^{\circ}=\{y\in\mathbb{R}^{n}\mid\langle x,y\rangle=\sum_{i=1}^{n}x_{i}y_{i}\leq 1\,\,\forall x\in K\} is the polar body of KK. While it is well known [Santaló 1949] that the volume product among convex bodies whose barycenter is centered at the origin is maximized only by ellipsoids, the lower bound is still unknown in dimension greater than 22. More precisely, the Mahler conjecture claims that any convex body KK containing the origin in its interior should satisfy the following optimal inequality:

vp(K)(n+1)n+1(n!)2.\operatorname{vp}(K)\geq\frac{(n+1)^{n+1}}{(n!)^{2}}.

The conjectured lower bound is reached by any nn-simplex whose vertices span n\mathbb{R}^{n} and sum up to zero. Mahler actually proved the inequality in dimension n=2n=2 [Mahler 1939], but so far the conjecture remains open for higher dimensions.

If we endow the space of convex bodies in n\mathbb{R}^{n} with the usual Hausdorff topology, the volume product becomes a continuous function over the space 𝒦0n\mathcal{K}_{0}^{n} of convex bodies in n\mathbb{R}^{n} containing the origin in their interior. A standard compactness argument then shows that the functional vp:𝒦0n(0,)\operatorname{vp}:\mathcal{K}_{0}^{n}\to(0,\infty) attains its minimum. According to [Santaló 1949] any local minimizer KK of the volume product satisfies that its barycenter 𝔾K\mathbb{G}_{K} and the barycenter 𝔾K\mathbb{G}_{K^{\circ}} of its polar coincide with the origin. Now recall that the covariance matrix cov(K)\operatorname{cov}(K) of a convex body KK whose barycenter is centered at the origin is the symmetric matrix with the following entries:

cov(K)ij=1|K|Kxixjdx.\operatorname{cov}(K)_{ij}=\frac{1}{|K|}\int_{K}x_{i}x_{j}dx.

In this paper we propose a short proof of the following result due to Klartag.

Theorem 1.1.

[Klartag 2018] Let KK be a local minimizer of the volume product. Then

cov(K)1(n+2)2cov(K)1\operatorname{cov}(K^{\circ})\geq\frac{1}{(n+2)^{2}}\cdot\operatorname{cov}(K)^{-1}

in the sense of symmetric matrices.

It follows from this result that a strong version of the slicing conjecture implies the Mahler conjecture. This strong slicing conjecture states that the homogeneous quantity det[cov(K)]/|K|2\det[\operatorname{cov}(K)]/|K|^{2} is maximized among convex bodies with barycenter at the origin by any nn-simplex whose vertices span n\mathbb{R}^{n} and sum up to zero, see [Klartag 2018] for further details.

Klartag’s proof of Theorem 1.1 is based on a variational argument in the space of projective images of KK that makes substantial use of the Laplace transform. In this paper we will also use a variational argument in the space of projective images of KK in order to recover Theorem 1.1, but our computation will be considerably simpler. More precisely, in section 2 we consider the functional K(x,ξ)vp((K+ξ)+x)\mathcal{F}_{K}(x,\xi)\coloneqq\operatorname{vp}\left((K^{\circ}+\xi)^{\circ}+x\right) which is well-defined for any convex body KK in 𝒦0n\mathcal{K}_{0}^{n} on an open neighborhood of the origin in n×n\mathbb{R}^{n}\times\mathbb{R}^{n}. By [Santaló 1949] we will easily observe (Proposition 2.1) that the origin is a critical point of the functional K\mathcal{F}_{K} if and only if 𝔾K=𝔾K=0\mathbb{G}_{K}=\mathbb{G}_{K^{\circ}}=0. Then an elementary computation (Proposition 2.2) will lead us to the following expression for the Hessian matrix at such a critical point:

Hess(0,0)K=(n+1)vp(K)((n+2)cov(K)InIn(n+2)cov(K)).\operatorname{Hess}_{(0,0)}\mathcal{F}_{K}=(n+1)\operatorname{vp}(K)\left(\begin{array}[]{cc}(n+2)\operatorname{cov}(K^{\circ})&-I_{n}\\ -I_{n}&(n+2)\operatorname{cov}(K)\end{array}\right).

Finally we will check (Proposition 2.3) that the above Hessian is positive semi-definite if and only if the symmetric matrix inequality (n+2)cov(K)[(n+2)cov(K)]1(n+2)\cdot\operatorname{cov}(K^{\circ})\geq[(n+2)\cdot\operatorname{cov}(K)]^{-1} holds true. This implies Theorem 1.1.

Our functional K\mathcal{F}_{K} is defined using special projective transformations of the convex body, which are obtained as specific combinations of translations of the polar body with translations of the body itself. In section 3 we will prove that this functional captures all the relevant information about variational properties of the volume product on the space of projective images of a given convex body. More precisely, we will first prove in Proposition 3.1 that once a suitable affine coordinate chart is fixed, the projective transformations close to the identity map transform the convex body KK into a new convex body which still contains the origin in its interior. We will then show in Proposition 3.2 that the volume product of these small projective perturbations of our convex body factors through the map K\mathcal{F}_{K}.

Finally in section 4 we will exhibit an interesting family of convex bodies in dimension 22 that are critical points of the volume product on the space of their own projective images. More precisely, we will show the following.

Theorem 1.2.

There exists a 11-parameter deformation {Pq}q[1/2,1/2]\{P_{q}\}_{q\in[-1/\sqrt{2},1/\sqrt{2}]} of an equilateral triangle Δ=P1/2\Delta=P_{-1/\sqrt{2}} by axially symmetric pentagons satisfying the following properties.

  1. i)

    Every pentagon PqP_{q} is a critical point of the volume product in its own projective orbit; more precisely, the function Pq\mathcal{F}_{P_{q}} has a critical point at the origin.

  2. ii)

    The pentagon P0P_{0} is a saddle point; more precisely the Hessian matrix Hess(0,0)P0\operatorname{Hess}_{(0,0)}\mathcal{F}_{P_{0}} is non-degenerate and non-definite.

  3. iii)

    For qq close enough to 1/2-1/\sqrt{2} we have

    (1.1) tr[cov(Pq)cov(Pq)]>tr[cov(Δ)cov(Δ)]=1/8.\operatorname{tr}[\operatorname{cov}(P_{q})\operatorname{cov}(P_{q}^{\circ})]>\operatorname{tr}[\operatorname{cov}(\Delta)\operatorname{cov}(\Delta^{\circ})]=1/8.

The main relevant property of this family is that part of these critical points are saddle points. The last property that there exists small deformations of Δ\Delta verifying (1.1) is also remarkable for the following reason. Recall that the 22-dimensional Euclidean unit ball BB also satisfies tr[cov(B)cov(B)]=1/8\operatorname{tr}[\operatorname{cov}(B)\operatorname{cov}(B^{\circ})]=1/8, and that Kuperberg in a private communication to Klartag shows that the Euclidean ball of any dimension is a local maximizer of the functional Ktr[cov(K)cov(K)]K\mapsto\operatorname{tr}[\operatorname{cov}(K)\operatorname{cov}(K^{\circ})] in the class of centrally-symmetric convex bodies with sufficiently smooth boundary, see [Klartag 2018, p.78]. At some point it was not clear if this upperbound was a global one, until it was disproved in [Klartag 2018, Proposition 1.5] for large dimensions. Our family discards any non-symmetric analog of Kuperberg’s local result near the nn-simplex, even in low dimensions.

Acknowledgements. We would like to thank B. Klartag for valuable exchanges.

2. A simple variational study of the volume product

The purpose of this section is to give a short proof of Theorem 1.1.

Fix a convex body KK of n\mathbb{R}^{n} containing the origin in its interior, and consider the functional

K:UK\displaystyle\mathcal{F}_{K}:\,\,U_{K} \displaystyle\to (0,)\displaystyle(0,\infty)
(x,ξ)\displaystyle(x,\xi) \displaystyle\mapsto vp((K+ξ)+x)\displaystyle\operatorname{vp}\left((K^{\circ}+\xi)^{\circ}+x\right)

which is well-defined in an open neighbourhood UKU_{K} of (0,0)n×n(0,0)\in\mathbb{R}^{n}\times\mathbb{R}^{n}.

Proposition 2.1.

The Jacobian matrix of K\mathcal{F}_{K} at the origin is the following:

Jac(0,0)K=(n+1)vp(K)(𝔾K𝔾K).\operatorname{Jac}_{(0,0)}\mathcal{F}_{K}=-(n+1)\operatorname{vp}(K)\left(\begin{array}[]{cc}\mathbb{G}_{K^{\circ}}&\mathbb{G}_{K}\end{array}\right).

Recall the barycenter of a convex body AA is defined by the formula 𝔾A|A|1Ax𝑑x\mathbb{G}_{A}\coloneqq|A|^{-1}\int_{A}xdx.

Proof.

Equivalently, we shall prove that for any (v,η)n×n(v,\eta)\in\mathbb{R}^{n}\times\mathbb{R}^{n}, the following holds:

d(0,0)K(v,η)=(n+1)vp(K)(𝔾K,v+𝔾K,η).d_{(0,0)}\,\mathcal{F}_{K}(v,\eta)=-(n+1)\operatorname{vp}(K)(\langle{\mathbb{G}}_{K^{\circ}},v\rangle+\langle{\mathbb{G}}_{K},\eta\rangle).

First recall that Santaló proved in [Santaló 1949] the following formula:

ddt|0|(Ktv)|=(n+1)|K|𝔾K,v.\left.\frac{d}{dt}\right|_{0}|(K-tv)^{\circ}|=(n+1)\cdot|K^{\circ}|\cdot\left\langle{\mathbb{G}}_{K^{\circ}},v\right\rangle.

Then we easily compute that

d(0,0)K(v,0)=ddt|0vp(K+tv)=(n+1)vp(K)𝔾K,v.d_{(0,0)}\,\mathcal{F}_{K}(v,0)=\left.\frac{d}{dt}\right|_{0}\operatorname{vp}(K+tv)=-(n+1)\cdot\operatorname{vp}(K)\cdot\left\langle{\mathbb{G}}_{K^{\circ}},v\right\rangle.

The equality d(0,0)K(0,η)=(n+1)vp(K)𝔾K,ηd_{(0,0)}\,\mathcal{F}_{K}(0,\eta)=-(n+1)\cdot\operatorname{vp}(K)\cdot\langle{\mathbb{G}}_{K},\eta\rangle easily follows from the previous computation using that vp(K)=vp(K)\operatorname{vp}(K)=\operatorname{vp}(K^{\circ}). ∎

In particular the origin is a critical point of the functional K\mathcal{F}_{K} if and only if 𝔾K=𝔾K=0\mathbb{G}_{K}=\mathbb{G}_{K^{\circ}}=0.

Proposition 2.2.

Suppose that the origin is a critical point of K\mathcal{F}_{K}. Then its Hessian matrix is the following:

Hess(0,0)K=(n+1)vp(K)((n+2)cov(K)InIn(n+2)cov(K)).\operatorname{Hess}_{(0,0)}\mathcal{F}_{K}=(n+1)\operatorname{vp}(K)\left(\begin{array}[]{cc}(n+2)\operatorname{cov}(K^{\circ})&-I_{n}\\ -I_{n}&(n+2)\operatorname{cov}(K)\end{array}\right).
Proof.

First observe that for any v,wnv,w\in\mathbb{R}^{n}

Hess(0,0)K[(v,0),(w,0)]\displaystyle\operatorname{Hess}_{(0,0)}\,\mathcal{F}_{K}[(v,0),(w,0)] =ddt|0dds|0K(tv+sw,0)\displaystyle=\left.\frac{d}{dt}\right|_{0}\left.\frac{d}{ds}\right|_{0}\mathcal{F}_{K}(tv+sw,0)
(using the previous proposition)\displaystyle(\text{using the previous proposition}) =(n+1)ddt|0vp(K+tv)𝔾(K+tv),w\displaystyle=-(n+1)\cdot\left.\frac{d}{dt}\right|_{0}\operatorname{vp}(K+tv)\cdot\left\langle{\mathbb{G}}_{(K+tv)^{\circ}},w\right\rangle
=(n+1)|K|ddt|0[|(K+tv)|𝔾(K+tv)],w.\displaystyle=-(n+1)\cdot|K|\cdot\left\langle\left.\frac{d}{dt}\right|_{0}[|(K+tv)^{\circ}|\cdot{\mathbb{G}}_{(K+tv)^{\circ}}],w\right\rangle.

Now, using the radial function ρA\rho_{A} of a convex body A𝒦0nA\in\mathcal{K}_{0}^{n}, the support function hAh_{A^{\circ}} of its polar set, and the fact that ρA=1/hA\rho_{A}=1/h_{A^{\circ}} we compute

ddt|0[|(K+tv)|𝔾(K+tv)]\displaystyle\left.\frac{d}{dt}\right|_{0}[|(K+tv)^{\circ}|\cdot{\mathbb{G}}_{(K+tv)^{\circ}}] =ddt|0(K+tv)x𝑑x\displaystyle=\left.\frac{d}{dt}\right|_{0}\int_{(K+tv)^{\circ}}xdx
=ddt|0Sn10ρ(K+tv)(u)(ru)rn1𝑑r𝑑u\displaystyle=\left.\frac{d}{dt}\right|_{0}\int_{S^{n-1}}\int_{0}^{\rho_{(K+tv)^{\circ}}(u)}(ru)\,r^{n-1}drdu
=1n+1ddt|0Sn1ρ(K+tv)(u)n+1u𝑑u\displaystyle=\frac{1}{n+1}\cdot\left.\frac{d}{dt}\right|_{0}\int_{S^{n-1}}\rho_{(K+tv)^{\circ}}(u)^{n+1}u\,du
=1n+1ddt|0Sn1uhK+tv(u)n+1𝑑u\displaystyle=\frac{1}{n+1}\cdot\left.\frac{d}{dt}\right|_{0}\int_{S^{n-1}}\frac{u}{h_{K+tv}(u)^{n+1}}du
=1n+1ddt|0Sn11(hK(u)+tv,u)n+1u𝑑u\displaystyle=\frac{1}{n+1}\cdot\left.\frac{d}{dt}\right|_{0}\int_{S^{n-1}}\frac{1}{(h_{K}(u)+t\langle v,u\rangle)^{n+1}}\,udu
=Sn1v,uhK(u)n+2u𝑑u.\displaystyle=-\int_{S^{n-1}}\frac{\langle v,u\rangle}{h_{K}(u)^{n+2}}\,udu.

Therefore

Hess(0,0)K((v,0),(w,0))\displaystyle\operatorname{Hess}_{(0,0)}\,\mathcal{F}_{K}((v,0),(w,0)) =(n+1)|K|Sn1v,uw,uhK(u)n+2𝑑u\displaystyle=(n+1)\cdot|K|\cdot\int_{S^{n-1}}\frac{\langle v,u\rangle\cdot\langle w,u\rangle}{h_{K}(u)^{n+2}}\,du
=(n+1)(n+2)|K|Kv,xw,x𝑑x\displaystyle=(n+1)\cdot(n+2)\cdot|K|\cdot\int_{K^{\circ}}\langle v,x\rangle\cdot\langle w,x\rangle\,dx
=(n+1)vp(K)(n+2)vtcov(K)w\displaystyle=(n+1)\cdot\operatorname{vp}(K)\cdot(n+2)\cdot v^{t}\operatorname{cov}(K^{\circ})\ w

as claimed. The computation of the dual term

Hess(0,0)K((0,ξ),(0,η))=(n+1)vp(K)(n+2)ξtcov(K)η.\operatorname{Hess}_{(0,0)}\,\mathcal{F}_{K}((0,\xi),(0,\eta))=(n+1)\cdot\operatorname{vp}(K)\cdot(n+2)\cdot\xi^{t}\operatorname{cov}(K)\ \eta.

for any ξ,ηn\xi,\eta\in\mathbb{R}^{n} is similar. It remains to compute the mixed term as follows:

Hess(0,0)K((v,0),(0,η))\displaystyle\operatorname{Hess}_{(0,0)}\,\mathcal{F}_{K}((v,0),(0,\eta)) =dds|0(|(K+sη)|ddt|0|((K+sη)+tv)|)\displaystyle=\left.\frac{d}{ds}\right|_{0}\left(|(K^{\circ}+s\eta)^{\circ}|\cdot\left.\frac{d}{dt}\right|_{0}|((K^{\circ}+s\eta)^{\circ}+tv)^{\circ}|\right)
=(n+1)dds|0(vp(K+sη)𝔾K+sη,v)\displaystyle=-(n+1)\left.\frac{d}{ds}\right|_{0}\left(\operatorname{vp}(K^{\circ}+s\eta)\cdot\langle{\mathbb{G}}_{K^{\circ}+s\eta},v\rangle\right)
(the origin being a critical point)\displaystyle(\text{the origin being a critical point}) =(n+1)vp(K)dds|0𝔾K+sη,v\displaystyle=-(n+1)\cdot\operatorname{vp}(K^{\circ})\cdot\left.\frac{d}{ds}\right|_{0}\langle{\mathbb{G}}_{K^{\circ}+s\eta},v\rangle
=(n+1)vp(K)η,v.\displaystyle=-(n+1)\cdot\operatorname{vp}(K)\cdot\langle\eta,v\rangle.

This completes the proof. ∎

Proposition 2.3.

Suppose that the origin is a critical point of K\mathcal{F}_{K}. Then

Hess(0,0)K0cov(K)1(n+2)2cov(K)1.\operatorname{Hess}_{(0,0)}\mathcal{F}_{K}\geq 0\Longleftrightarrow\operatorname{cov}(K^{\circ})\geq\frac{1}{(n+2)^{2}}\cdot\operatorname{cov}(K)^{-1}.

Moreover

(2.1) det(Hess(0,0)K)=(n+1)nvp(K)ndet[(n+2)2cov(K)cov(K)In].\det(\operatorname{Hess}_{(0,0)}\mathcal{F}_{K})=(n+1)^{n}\operatorname{vp}(K)^{n}\det[(n+2)^{2}\operatorname{cov}(K^{\circ})\operatorname{cov}(K)-I_{n}].

As observed in the introduction, the first part of this statement implies Theorem 1.1. The second part will be useful in the last section.

Proof.

Just remark that if AA and BB are two symmetric positive definite matrices, and if we set

M(AInInB)andP(InA10In),M\coloneqq\left(\begin{array}[]{cc}A&-I_{n}\\ -I_{n}&B\end{array}\right)\ \ \text{and}\ \ P\coloneqq\left(\begin{array}[]{cc}I_{n}&A^{-1}\\ 0&I_{n}\end{array}\right),

then PP being invertible the following holds true:

M0PtMP=(A00BA1)0BA1.M\geq 0\ \ \Longleftrightarrow P^{t}\ M\ P=\left(\begin{array}[]{cc}A&0\\ 0&B-A^{-1}\end{array}\right)\geq 0\ \Longleftrightarrow B\geq A^{-1}.

Since detP=1\det P=1,

detM=det(A00BA1)=det(A)det(BA1)=det(ABIn).\det M=\det\left(\begin{array}[]{cc}A&0\\ 0&B-A^{-1}\end{array}\right)=\det(A)\det(B-A^{-1})=\det(AB-I_{n}).\qed

3. Projective transformations and volume product

The purpose of this section is to show that the volume product on the space of projective images of a given convex body KK factors through the map K\mathcal{F}_{K}. Therefore the functional K\mathcal{F}_{K} captures all the relevant information about variational properties of the volume product on the space of projective images of a given convex body.

First recall some material about projective transformations (also called homographies). Consider the real projective space n=(n+1{0})/\mathbb{RP}^{n}=(\mathbb{R}^{n+1}\setminus\{0\})/\mathbb{R}^{*}. We will make use of the projective coordinates [x0,,xn][x_{0},\ldots,x_{n}] for points in n\mathbb{RP}^{n}. Recall that a projective transformation is a diffeomorphism of n\mathbb{RP}^{n} induced by a linear isomorphism of n+1\mathbb{R}^{n+1} as follows: given A=(aij)GLn+1()A=(a_{ij})\in\operatorname{GL}_{n+1}(\mathbb{R}) the induced projective transformation A\mathbb{P}{A} is

A[x0,,xn]=[i=0na0ixi,,i=0nanixi].\mathbb{P}{A}[x_{0},\ldots,x_{n}]=\left[\sum_{i=0}^{n}a_{0i}x_{i},\ldots,\sum_{i=0}^{n}a_{ni}x_{i}\right].

This allows to identify the group of projective transformations Proj(n)\operatorname{Proj}(\mathbb{RP}^{n}) with the projective linear group PGLn()=GLn+1()/(In+1)\operatorname{PGL}_{n}(\mathbb{R})=\operatorname{GL}_{n+1}(\mathbb{R})/(\mathbb{R}^{\ast}\cdot I_{n+1}) and defines a natural topology on it. Here In+1I_{n+1} denotes the identity matrix. We will use the affine coordinate chart

F:n\displaystyle F:\mathbb{R}^{n} \displaystyle\longrightarrow nH0\displaystyle\mathbb{RP}^{n}\setminus H_{0}
y=(y1,,yn)\displaystyle y=(y_{1},\ldots,y_{n}) \displaystyle\longmapsto [1,y]=[1,y1,,yn],\displaystyle[1,y]=[1,y_{1},\ldots,y_{n}],

where H0={[x0,x1,,xn]x0=0}H_{0}=\{[x_{0},x_{1},\ldots,x_{n}]\mid x_{0}=0\}. Its inverse map is F1[x0,,xn]=(x1/x0,,xn/x0)F^{-1}[x_{0},\ldots,x_{n}]=(x_{1}/x_{0},\ldots,x_{n}/x_{0}).

Given two nested open balls centered at the origin B1B2B_{1}\varsubsetneq B_{2}, we define the open subset of 𝒦0n\mathcal{K}_{0}^{n}

𝒱{K𝒦0ncl(B1)int(K)KB2}\mathcal{V}\coloneqq\{K\in\mathcal{K}_{0}^{n}\mid\operatorname{cl}({B}_{1})\subset\operatorname{int}(K)\subset K\subset B_{2}\}

where cl\operatorname{cl} and int\operatorname{int} refer to the closure and the interior respectively.

Proposition 3.1.

There exists an open neighbourhood 𝒰\mathcal{U} of the identity in Proj(n)\operatorname{Proj}(\mathbb{RP}^{n}) such that K𝒱\forall K\in\mathcal{V} and π𝒰\forall\pi\in\mathcal{U} we have πF(K)nH0\pi\circ F(K)\subset\mathbb{RP}^{n}\setminus H_{0} and F1πF(K)𝒦0nF^{-1}\circ\pi\circ F(K)\in\mathcal{K}_{0}^{n}.

Proof.

Let us introduce the following local parametrization

Φ:GLn()×n×n\displaystyle\Phi:\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n} \displaystyle\to Proj(n)PGLn()\displaystyle\operatorname{Proj}(\mathbb{RP}^{n})\simeq\operatorname{PGL}_{n}(\mathbb{R})
(M,x,ξ)\displaystyle(M,x,\xi) \displaystyle\mapsto (1ξtxM).\displaystyle\mathbb{P}\begin{pmatrix}1&\xi^{t}\\ x&M\end{pmatrix}.

Clearly Φ\Phi is a diffeomorphism onto its image which is an open neighbourhood of the identity. We easily compute for any π=Φ(M,x,ξ)\pi=\Phi(M,x,\xi) and every yny\in\mathbb{R}^{n} that

πF(y)=[1+ξty,x+My].\pi\circ F(y)=[1+\xi^{t}y,x+My].

So πF(y)nH0\pi\circ F(y)\in\mathbb{RP}^{n}\setminus H_{0} if and only if 1+ξty01+\xi^{t}y\neq 0, and the condition that πF(K)nH0\pi\circ F(K)\subset\mathbb{RP}^{n}\setminus H_{0} amounts to the condition that ξint(K)-\xi\in\operatorname{int}(K^{\circ}). Now if ξint(K)-\xi\in\operatorname{int}(K^{\circ}) the well-defined set F1πF(K)F^{-1}\circ\pi\circ F(K) is automatically a convex body, that contains the zero in its interior if and only if xint(MK)-x\in\operatorname{int}(MK). Therefore the open set 𝒰Φ({(M,x,ξ)ξB2andxMB1})\mathcal{U}\coloneqq\Phi\left(\{(M,x,\xi)\mid-\xi\in B_{2}^{\circ}\,\,\text{and}\,\,-x\in MB_{1}\}\right) satisfies the conclusion of the proposition. ∎

Now we will reparametrize this open neighbourhood 𝒰\mathcal{U} in order to show that the volume product of projective images of KK factors through the map K\mathcal{F}_{K}.

Proposition 3.2.

There exists an open neighbourhood 𝒲\mathcal{W} of (In,0,0)(I_{n},0,0) in GLn()×n×n\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n} and a map Ψ:𝒲Proj(n)\Psi:\mathcal{W}\to\operatorname{Proj}(\mathbb{RP}^{n}) such that Ψ(In,0,0)=Idn\Psi(I_{n},0,0)=\operatorname{Id}_{\mathbb{RP}^{n}}, Ψ\Psi is a diffeomorphism onto its image, and the following holds:

vp(F1Ψ(M,x,ξ)F(K))=K(x,ξ)\operatorname{vp}(F^{-1}\circ\Psi(M,x,\xi)\circ F(K))=\mathcal{F}_{K}(x,\xi)

for any convex body K𝒱K\in\mathcal{V} and for any (M,x,ξ)Ψ1(𝒰)(M,x,\xi)\in\Psi^{-1}(\mathcal{U}).

Therefore, in order to study the variational properties of the volume product of projective images of a convex body K𝒦0nK\in\mathcal{K}_{0}^{n}, it is enough to study the associated functional K\mathcal{F}_{K}.

Proof.

Consider the map

α:GLn()×n×n\displaystyle\alpha:\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n} \displaystyle\to GLn()×n×n\displaystyle\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n}
(N,y,η)\displaystyle(N,y,\eta) \displaystyle\mapsto (N(In+yηt),Ny,η).\displaystyle\left(N(I_{n}+y\eta^{t}),Ny,\eta\right).

The map α\alpha fixes points of the form (M,0,0)(M,0,0) for any MGLn()M\in\operatorname{GL}_{n}(\mathbb{R}), and there exists an open neighbourhood 𝒲\mathcal{W} of (In,0,0)(I_{n},0,0) in GLn()×n×n\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n} such that the restriction of α\alpha to 𝒲\mathcal{W} is a diffeomorphism onto its image whose inverse is α1(M,x,ξ)=(Mxξt,(Mxξt)1x,ξ)\alpha^{-1}(M,x,\xi)=\Big{(}M-x\xi^{t},(M-x\xi^{t})^{-1}x,\xi\Big{)}. Then define the map

Ψ:GLn()×n×n\displaystyle\Psi:\operatorname{GL}_{n}(\mathbb{R})\times\mathbb{R}^{n}\times\mathbb{R}^{n} \displaystyle\to Proj(n)\displaystyle\operatorname{Proj}(\mathbb{RP}^{n})
(N,y,η)\displaystyle(N,y,\eta) \displaystyle\mapsto (100N)(10yIn)(1ηt0In).\displaystyle\mathbb{P}\begin{pmatrix}1&0\\ 0&N\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&0\\ y&I_{n}\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&\eta^{t}\\ 0&I_{n}\end{pmatrix}.

Because

Φα(N,y,η)=(1ηtNyN(In+yηt)),\Phi\circ\alpha(N,y,\eta)=\mathbb{P}\begin{pmatrix}1&\eta^{t}\\ Ny&N(I_{n}+y\eta^{t})\end{pmatrix},

while

(100N)(10yIn)(1ηt0In)=(10NyN)(1ηt0In)=(1ηtNyN(In+yηt)),\mathbb{P}\begin{pmatrix}1&0\\ 0&N\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&0\\ y&I_{n}\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&\eta^{t}\\ 0&I_{n}\end{pmatrix}=\mathbb{P}\begin{pmatrix}1&0\\ Ny&N\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&\eta^{t}\\ 0&I_{n}\end{pmatrix}=\mathbb{P}\begin{pmatrix}1&\eta^{t}\\ Ny&N(I_{n}+y\eta^{t})\end{pmatrix},

we have that Ψ=Φα\Psi=\Phi\circ\alpha which implies that the restriction of Ψ\Psi to 𝒲\mathcal{W} is a diffeomorphism onto its image. By construction Ψ(In,0,0)=Idn\Psi(I_{n},0,0)=\operatorname{Id}_{\mathbb{RP}^{n}}. Still denote by Ψ:𝒲Proj(n)\Psi:\mathcal{W}\to\operatorname{Proj}(\mathbb{RP}^{n}) the restriction of the map Ψ\Psi to 𝒲\mathcal{W}. Now for any convex body K𝒱K\in\mathcal{V} and for any (M,x,ξ)Ψ1(𝒰)(M,x,\xi)\in\Psi^{-1}(\mathcal{U}) the following holds:

F1Ψ(M,x,ξ)F(K)\displaystyle F^{-1}\circ\Psi(M,x,\xi)\circ F(K) =F1(100M)(10xIn)(1ξt0In)F(K)\displaystyle=F^{-1}\circ\mathbb{P}\begin{pmatrix}1&0\\ 0&M\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&0\\ x&I_{n}\end{pmatrix}\cdot\mathbb{P}\begin{pmatrix}1&\xi^{t}\\ 0&I_{n}\end{pmatrix}\circ F(K)
=F1(100M)Ψ(In,x,ξ)F(K)\displaystyle=F^{-1}\circ\mathbb{P}\begin{pmatrix}1&0\\ 0&M\end{pmatrix}\cdot\Psi(I_{n},x,\xi)\circ F(K)
=F1(100M)FF1Ψ(In,x,ξ)F(K).\displaystyle=F^{-1}\circ\mathbb{P}\begin{pmatrix}1&0\\ 0&M\end{pmatrix}\circ F\circ F^{-1}\circ\Psi(I_{n},x,\xi)\circ F(K).

But for all yny\in\mathbb{R}^{n}

F1(100M)F(y)=My,F^{-1}\circ\mathbb{P}\begin{pmatrix}1&0\\ 0&M\end{pmatrix}\circ F(y)=My,

which implies using the invariance of the volume product by the linear group that

vp(F1Ψ(M,x,ξ)F(K))=vp(F1Ψ(In,x,ξ)F(K)).\operatorname{vp}(F^{-1}\circ\Psi(M,x,\xi)\circ F(K))=\operatorname{vp}(F^{-1}\circ\Psi(I_{n},x,\xi)\circ F(K)).

Now observe that

F1Ψ(In,x,ξ)F(K)=(K+ξ)+x.F^{-1}\circ\Psi(I_{n},x,\xi)\circ F(K)=(K^{\circ}+\xi)^{\circ}+x.

Indeed write F1Ψ(In,x,ξ)F=F1Ψ(In,x,0)FF1Ψ(In,0,ξ)FF^{-1}\circ\Psi(I_{n},x,\xi)\circ F=F^{-1}\circ\Psi(I_{n},x,0)\circ F\circ F^{-1}\circ\Psi(I_{n},0,\xi)\circ F. First F1Ψ(In,x,0)F(y)=x+yF^{-1}\circ\Psi(I_{n},x,0)\circ F(y)=x+y while F1Ψ(In,0,ξ)F(y)=y/(1+ξty)F^{-1}\circ\Psi(I_{n},0,\xi)\circ F(y)=y/(1+\xi^{t}y). By a straightforward computation it implies that F1Ψ(In,0,ξ)F(K)=(K+ξ)F^{-1}\circ\Psi(I_{n},0,\xi)\circ F(K)=(K^{\circ}+\xi)^{\circ}.

It follows that vp(F1Ψ(M,x,ξ)F(K))=vp((K+ξ)+x)\operatorname{vp}(F^{-1}\circ\Psi(M,x,\xi)\circ F(K))=\operatorname{vp}\left((K^{\circ}+\xi)^{\circ}+x\right) as claimed. ∎

4. A family of critical pentagons

Fix two numbers b>0b>0 and q[1b,b]q\in[-\frac{1}{b},b], and set r=1+q2r=\sqrt{1+q^{2}} and c=1+bq1+q2c=\frac{1+bq}{\sqrt{1+q^{2}}}.

We consider the pentagon Pq,b2P_{q,b}\subset\mathbb{R}^{2} whose vertices are (1b,0)(-\frac{1}{b},0), (q,±r)(q,\pm r) and (b,±c)(b,\pm c). By construction P=Pq,bP=P_{q,b} is symmetric with respect to the xx-axis:

[Uncaptioned image]
Lemma 4.1.

The dual of PP is P-P.

Proof.

This is a direct consequence of the following general fact for convex polygons: a point (A,B)(A,B) is a vertex of PP^{\circ} if and only if the line Ax+By=1Ax+By=1 contains a side of PP. In this way:

  • the line through (1q,0)(-\frac{1}{q},0) and (0,±1r)(0,\pm\frac{1}{r}) containing a side of PP corresponds to the vertex (q,±r)(-q,\pm r) of PP^{\circ},

  • the line through (1b,0)(-\frac{1}{b},0) and (0,±1c)(0,\pm\frac{1}{c}) containing a side of PP corresponds to the vertex (b,±c)(-b,\pm c) of PP^{\circ},

  • the line through (b,c)(b,c) and (b,c)(b,-c) containing a side of PP corresponds to the vertex (1b,0)(\frac{1}{b},0) of PP^{\circ}.

Therefore PP^{\circ} coincides with P-P. ∎

By the previous lemma, if the barycenter of PP lies at the origin, the same holds for its dual P-P, and thus PP is a critical point for the volume product on the space of projective images of PP.

The next lemma gives the exact condition on the pair (q,b)(q,b) which ensures that PP is such a critical point.

Lemma 4.2.

The barycenter of PP lies at the origin if and only if

f(q,b)q3bq2b2q2+2qb5+qb+3b4=1.f(q,b)\coloneqq q^{3}b-q^{2}b^{2}-q^{2}+2qb^{5}+qb+3b^{4}=1.
Proof.

Since PP is symmetric, its barycenter lies on the xx-axis, and therefore only the abscissa of the barycenter is relevant. This coordinate can be computed by decomposing the pentagon into six pairwise symmetrical triangles, with one vertex at the origin and one common side with PP. The condition on the barycenter then becomes

(b+b)3|bb0c|+(b+q)3|bqcr|+(q1b)3|q1br0|=0.\frac{(b+b)}{3}\begin{vmatrix}b&b\\ 0&c\end{vmatrix}+\frac{(b+q)}{3}\begin{vmatrix}b&q\\ c&r\end{vmatrix}+\frac{(q-\frac{1}{b})}{3}\begin{vmatrix}q&-\frac{1}{b}\\ r&0\end{vmatrix}=0.

Multiplying by 3b2r3b^{2}r and using that r2=1+q2r^{2}=1+q^{2} and rc=1+bqrc=1+bq gives the desired condition. ∎

As concrete examples of pairs (q,b)(q,b) such that PP is critical, i.e. solutions of the equation in Lemma 4.2, we can consider the following pairs:

  • the two pairs (1/2,2)(-1/\sqrt{2},\sqrt{2}) and (1/2,1/2)(1/\sqrt{2},1/\sqrt{2}) at which the pentagon degenerates respectively into an equilateral triangle Δ\Delta and its dual Δ-\Delta,

  • the pair (cos(2π/5)/cos(π/5),cos(π/5))(-\cos(2\pi/5)/\sqrt{\cos(\pi/5)},\sqrt{\cos(\pi/5)}) at which the pentagon is regular,

  • the remarkable pair (0,1/34)(0,1/\sqrt[4]{3}) which corresponds to a critical saddle point as we will show later.

In addition to these few simple concrete examples, the next lemma 4.3 shows that we have indeed a whole continuous family of critical pentagons {Pq}q[1/2,1/2]\{P_{q}\}_{q\in[-1/\sqrt{2},1/\sqrt{2}]}.

Lemma 4.3.

The set

{(q,b)[12,12]×[12,2]|q3bq2b2q2+2qb5+qb+3b4=1}\biggl{\{}(q,b)\in[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\times[\frac{1}{\sqrt{2}},\sqrt{2}]\biggm{|}q^{3}b-q^{2}b^{2}-q^{2}+2qb^{5}+qb+3b^{4}=1\biggr{\}}

is the graph of a decreasing smooth function b(q)b(q) in the variable qq.

[Uncaptioned image]

As mentioned above, the extremal pentagons P±1/2P_{\pm 1/\sqrt{2}} degenerate to equilateral triangles.

Proof.

According to the implicit function theorem, the result follows from elementary computations that both partial derivatives of the function f(q,b)=q3bq2b2q2+2qb5+qb+3b4f(q,b)=q^{3}b-q^{2}b^{2}-q^{2}+2qb^{5}+qb+3b^{4} are strictly positive on this domain, except at point (1/2,1/2)(1/\sqrt{2},1/\sqrt{2}) where the first partial derivative is zero.

Indeed first observe that qf=3bq22(b2+1)q+2b5+b\partial_{q}f=3bq^{2}-2(b^{2}+1)q+2b^{5}+b is a quadratic polynomial in the variable qq. One can check that its discriminant 6b6+b4b2+1-6b^{6}+b^{4}-b^{2}+1 is always negative for b(1/2,2]b\in(1/\sqrt{2},\sqrt{2}], and zero at 1/21/\sqrt{2}, by studying its first and second derivative.

Next we compute that bf=q32bq2+(10b4+1)q+12b3\partial_{b}f=q^{3}-2bq^{2}+(10b^{4}+1)q+12b^{3} and check through the study of bqf\partial_{bq}f that bf(q,b)h(b)bf(1/2,b)\partial_{b}f(q,b)\geq h(b)\coloneqq\partial_{b}f(-1/\sqrt{2},b) which is a polynomial of degree 44. A straightforward variational study of hh ensures its positivity on the desired domain. ∎

Now we explain how to compute the Hessian of the functional P\mathcal{F}_{P} associated to some pentagon PP belonging the above family {Pq}\{P_{q}\}. For this, we must compute the inertia matrix II(P)=Pxixj𝑑xI\coloneqq I(P)=\int_{P}x_{i}x_{j}dx and the area |P||P| of such a pentagon. Since these quantities are additive with respect to disjoin union, we can perform these computations by decomposing the pentagon into triangles and using the following lemma.

Lemma 4.4.

The inertia matrix of a triangle Δ\Delta with vertices (0,0)(0,0), (x1,y1)(x_{1},y_{1}) and (x2,y2)(x_{2},y_{2}) is

I(Δ)=D12(Ixx(Δ)Ixy(Δ)Ixy(Δ)Iyy(Δ))I(\Delta)=\frac{D}{12}\begin{pmatrix}I_{xx}(\Delta)&I_{xy}(\Delta)\\ I_{xy}(\Delta)&I_{yy}(\Delta)\end{pmatrix}

where

{Ixx(Δ)=x12+x1x2+x22Ixy(Δ)=x1y1+x1y2+x2y12+x2y2Iyy(Δ)=y12+y1y2+y22\left\{\begin{aligned} I_{xx}(\Delta)&=x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}\\ I_{xy}(\Delta)&=x_{1}y_{1}+\frac{x_{1}y_{2}+x_{2}y_{1}}{2}+x_{2}y_{2}\\ I_{yy}(\Delta)&=y_{1}^{2}+y_{1}y_{2}+y_{2}^{2}\end{aligned}\right.

and D=|x1x2y1y2|=|x1y2x2y1|D=\begin{vmatrix}x_{1}&x_{2}\\ y_{1}&y_{2}\end{vmatrix}=|x_{1}y_{2}-x_{2}y_{1}| is twice the area of the triangle.

Proof.

Since we will not need the terms Ixy(Δ)I_{xy}(\Delta) in what follows, we will only do the computations for Ixx(Δ)I_{xx}(\Delta), the case of Iyy(Δ)I_{yy}(\Delta) being identical. An easy change of variables obtained by parametrizing our triangle by the triangle with vertices (0,0)(0,0), (1,0)(1,0) and (0,1)(0,1) gives us

Ixx(Δ)=0101t(sx1+tx2)2D𝑑s𝑑t=D12[x12+x1x2+x22].I_{xx}(\Delta)=\int_{0}^{1}\int_{0}^{1-t}(sx_{1}+tx_{2})^{2}Ddsdt=\frac{D}{12}\left[x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}\right].

By symmetry, the inertia matrix of such critical pentagon is diagonal so we only need to compute the terms Ixx(P)I_{xx}(P) and Iyy(P)I_{yy}(P). By decomposing as before PqP_{q} in six pairewise symmetrical triangles we get first that

(4.1) |Pq|=|qb1r0|+|bqcr|+|bb0c|.|P_{q}|=\begin{vmatrix}q&-b^{-1}\\ r&0\end{vmatrix}+\begin{vmatrix}b&q\\ c&r\end{vmatrix}+\begin{vmatrix}b&b\\ 0&c\end{vmatrix}.

Using the same decomposition and Lemma 4.4 we derive that

(4.2) Ixx(P)\displaystyle I_{xx}(P) =\displaystyle= 16|qb1r0|(b2+b1q+q2)\displaystyle\frac{1}{6}\begin{vmatrix}q&-b^{-1}\\ r&0\end{vmatrix}(-b^{-2}+-b^{-1}q+q^{2}) +16|bqcr|(q2+qb+b2)\displaystyle+\frac{1}{6}\begin{vmatrix}b&q\\ c&r\end{vmatrix}(q^{2}+qb+b^{2}) +\displaystyle+ 16|bb0c|(3b2)\displaystyle\frac{1}{6}\begin{vmatrix}b&b\\ 0&c\end{vmatrix}(3b^{2})
(4.3) Iyy(P)\displaystyle I_{yy}(P) =\displaystyle= 16|qb1r0|(r2)\displaystyle\frac{1}{6}\begin{vmatrix}q&-b^{-1}\\ r&0\end{vmatrix}(r^{2}) +16|bqcr|(r2+rc+c2)\displaystyle+\frac{1}{6}\begin{vmatrix}b&q\\ c&r\end{vmatrix}(r^{2}+rc+c^{2}) +\displaystyle+ 16|bb0c|(c2)\displaystyle\frac{1}{6}\begin{vmatrix}b&b\\ 0&c\end{vmatrix}(c^{2})

Let us define the functions

s(q,b)det(|P|2I216I(P)2)=19det(Hess(0,0)P),s(q,b)\coloneqq\det\left(|P|^{2}I_{2}-16I(P)^{2}\right)=\frac{1}{9}\det(\operatorname{Hess}_{(0,0)}\mathcal{F}_{P}),

where the second equality follows from equation (2.1), and

t(q,b)12tr(|P|2I216I(P)2)=8|P|2(tr(cov(Δ)cov(Δ))tr(cov(P)cov(P))).t(q,b)\coloneqq\frac{1}{2}\operatorname{tr}\left(|P|^{2}I_{2}-16I(P)^{2}\right)=8|P|^{2}\left(\operatorname{tr}(\operatorname{cov}(\Delta^{\circ})\operatorname{cov}(\Delta))-\operatorname{tr}(\operatorname{cov}(P^{\circ})\operatorname{cov}(P))\right).

For the pair (q,b)=(0,134)(q,b)=(0,-\frac{1}{\sqrt[4]{3}}), we get using (4.1),(4.2) and (4.3)

|P|=34(1+233),Ixx(P)=34(29+163),Iyy(P)=34(16+293)|P|=\sqrt[4]{3}\Big{(}1+\frac{2}{3}\sqrt{3}\Big{)},\qquad I_{xx}(P)=\sqrt[4]{3}\Big{(}\frac{2}{9}+\frac{1}{6}\sqrt{3}\Big{)},\qquad I_{yy}(P)=\sqrt[4]{3}\Big{(}\frac{1}{6}+\frac{2}{9}\sqrt{3}\Big{)}

and therefore

s(0,134)=1729(123+17)(4313)<0.s(0,-\frac{1}{\sqrt[4]{3}})=\frac{1}{729}(12\sqrt{3}+17)(4\sqrt{3}-13)<0.

Hence det(HessP0)\det(\operatorname{Hess}\mathcal{F}_{P_{0}}) is negative, and thus P0\mathcal{F}_{P_{0}} has a critical saddle point at the origin. This proves item ii) of Theorem 1.2.

To prove item iii) of Theorem 1.2, it will be enough to compute the derivative of t(q,b(q))t(q,b(q)) at q0=12q_{0}=-\frac{1}{\sqrt{2}}. First, at (q0,b0)=(12,2)(q_{0},b_{0})=(-\frac{1}{\sqrt{2}},\sqrt{2}) we have

|Pq|\displaystyle|P_{q}| =332,\displaystyle=\frac{3\sqrt{3}}{2}, q|Pq|=62,\displaystyle\frac{\partial}{\partial q}|P_{q}|=\frac{\sqrt{6}}{2}, b|Pq|=64,\displaystyle\frac{\partial}{\partial b}|P_{q}|=-\frac{\sqrt{6}}{4},
Ixx(Pq)\displaystyle I_{xx}(P_{q}) =338,\displaystyle=\frac{3\sqrt{3}}{8}, qIxx(Pq)=62,\displaystyle\frac{\partial}{\partial q}I_{xx}(P_{q})=\frac{\sqrt{6}}{2}, bIxx(Pq)=64,\displaystyle\frac{\partial}{\partial b}I_{xx}(P_{q})=-\frac{\sqrt{6}}{4},
Iyy(Pq)\displaystyle I_{yy}(P_{q}) =338,\displaystyle=\frac{3\sqrt{3}}{8}, qIyy(Pq)=68,\displaystyle\frac{\partial}{\partial q}I_{yy}(P_{q})=-\frac{\sqrt{6}}{8}, bIyy(Pq)=616.\displaystyle\frac{\partial}{\partial b}I_{yy}(P_{q})=-\frac{\sqrt{6}}{16}.

On the other hand, one has f(q0,b0)=924(6,1)\nabla f(q_{0},b_{0})=\frac{9\sqrt{2}}{4}(6,1) and thus bq(q0)=6\frac{\partial b}{\partial q}(q_{0})=-6. Therefore, using the previous values,

tq|(q0,b0)=924,tb|(q0,b0)=928,ddq|q0t(q,b(q))=tq+bqtb|(q0,b0)=92.\left.\frac{\partial t}{\partial q}\right|_{(q_{0},b_{0})}=-\frac{9\sqrt{2}}{4},\qquad\left.\frac{\partial t}{\partial b}\right|_{(q_{0},b_{0})}=\frac{9\sqrt{2}}{8},\qquad\left.\frac{d}{dq}\right|_{q_{0}}t(q,b(q))=\left.\frac{\partial t}{\partial q}+\frac{\partial b}{\partial q}\frac{\partial t}{\partial b}\right|_{(q_{0},b_{0})}=-{9\sqrt{2}}.

Since t(q0,b0)=0t(q_{0},b_{0})=0, this shows that t(q,b(q))<0t(q,b(q))<0 near q0q_{0}, thus finishing the proof of Theorem 1.2.

[Uncaptioned image]

To complete the picture, we conclude with a numerical analysis of the whole family of critical pentagons {Pq:q[1/2,1/2]}\{P_{q}\colon{q\in[-1/\sqrt{2},1/\sqrt{2}]}\}.

The previous three lemmas (4.1, 4.2 and 4.3) tell us that for any q[1/2,1/2]q\in[-1/\sqrt{2},1/\sqrt{2}] we can compute with very high precision (at least 101510^{-15} using python with scipy) the unique b[1/2,2]b\in[1/\sqrt{2},\sqrt{2}] such that Pq,bP_{q,b} is critical. Then all vertices are computed with precision at least 101410^{-14}, and therefore the polynomial function |P|28tr(I(P)2)|P|^{2}-8\operatorname{tr}\left(I(P)^{2}\right) is calculated with precision at least 101310^{-13} to obtain the above curve, which is negative if tr[cov(Pq)cov(Pq)]>1/8\operatorname{tr}[\operatorname{cov}(P_{q})\operatorname{cov}(P_{q}^{\circ})]>1/8 as cov(Pq)=I(Pq)/|Pq|\operatorname{cov}(P_{q})=I(P_{q})/|P_{q}|.

References

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