Subsection 3.4 (00TJ) · ScalingStacks

Verified tagged author-source HTML · 1912.02360v1 · cited publication edition alignment unverified.

3.4 Extension property and locally convex functions

We now discuss the issue of finding a tropical notion analogous to Kähler metrics. The concept of a Kähler metric is formulated in terms of a collection of local psh functions $\phi_{j}$ on overlapping complex charts, whose differences $\{\phi_{i}-\phi_{j}\}$ represent a given cocycle of local pluriharmonic function. Intuitively, the analogue should be a collection of local convex functions $u_{j}$ whose differences $\{u_{i}-u_{j}\}$ represent a given cocycle of local affine functions.

To the author’s awareness there is no definitive formulation of local convexity on polyhedral sets. In the case of interest, we need to define a class of ‘locally convex functions’ on $\partial\Delta_{\lambda}^{\vee}$ . The problem is that on $\widetilde{Sing}\subset\partial\Delta_{\lambda}^{\vee}$ , the transition functions between different charts are only piecewise linear, so convexity is not invariantly defined. This problem also prevents us from setting up a general global notion of real MA equation on $\partial\Delta_{\lambda}^{\vee}$ , which is an essential ingredient in the SYZ conjecture in general. We will attempt to give a special definition in the Fermat case (cf. section 3.5).

However, the extension theorem 2.13 provides an alternative viewpoint: (1,1)-type Kähler currents can be defined extrinsically. By analogy, we propose that the correct notion should be equivalent to the following

00QR

Definition 3.17. A continuous function $u$ on $\partial\Delta_{\lambda}^{\vee}$ satisfies the extension property if it extends to an admissible convex function on $N_{\mathbb{R}}=\mathbb{R}^{n+1}$ defined in section 3.3.

00QS

Example 3.18. The zero function extends to $L_{\lambda}$ , which is admissible and convex.

The problem is to make this definition both intrinsic to $\partial\Delta_{\lambda}^{\vee}$ , and local in nature. We do not fully succeed but shall make some partial progress.

00QT

Proposition 3.19. A continuous function $u$ on $\partial\Delta_{\lambda}^{\vee}$ satisfies the extension property if and only if for every $x\in\partial\Delta_{\lambda}^{\vee}$ , there exists $p\in\Delta$ , such that for any $y\in\partial\Delta_{\lambda}^{\vee}$ ,

u(y)\geq u(x)+\langle p,y-x\rangle.

00QU

Proof. The if direction is because the asymptotic growth condition (18) implies the gradient of $u$ must be contained in $\Delta$ .

For the only if direction, we apply the Legendre transform:

u^{*}(p)=\sup_{x\in\partial\Delta_{\lambda}^{\vee}}\{\langle x,p\rangle-u(x)\},\quad p\in\Delta,

and consider a version of the double Legendre transform

u^{**}(x)=\sup_{p\in\Delta}\{\langle x,p\rangle-u^{*}(p)\}.

Clearly $u^{**}$ is convex, and admissible by the boundedness of $u^{*}$ , and $u^{**}(x)\leq u(x)$ on $\partial\Delta_{\lambda}^{\vee}$ because

\langle x,p\rangle-u^{*}(p)\leq u(x),\quad\forall p\in\Delta.

Our characterisation precisely ensures $u^{**}(x)\geq u(x)$ on $\partial\Delta_{\lambda}^{\vee}$ . Then $u^{**}$ provides the canonical extension. ∎

00QV

Remark 3.20. The above characterisation is not completely intrinsic because it uses the extrinsic pairing $\langle,\rangle:M_{\mathbb{R}}\times N_{\mathbb{R}}\to\mathbb{R}$ . On the positive side it uses only the value of $u$ on $\partial\Delta_{\lambda}^{\vee}$ .

00QW

Remark 3.21. At $x\in\partial\Delta_{\lambda}^{\vee}$ , the vector $p\in\Delta$ in the hypothesis is a subgradient of the canonical extension $u^{**}$ , namely $u^{**}(y)-u(x)\geq\langle p,y-x\rangle$ .

We now introduce a local notion. The function $u$ below will be analogous to $\phi_{0}$ in (19). Recall the charts $\partial\Delta_{\lambda}^{\vee}\cap U_{w}^{\infty}$ associated to ourward normal vectors $w$ introduced in section 3.2, with local coordinates $x^{m_{1}},\ldots x^{m_{n}}$ .

00QX

Definition 3.22. Let $u$ be a continuous function on $\partial\Delta^{\vee}_{\lambda}$ , to which we associate a collection of local functions $\{u_{m}\}_{m\in\Delta_{\mathbb{Z}}}$ by the rule $u_{m}=u-\langle x,m\rangle$ . We regard $u_{m}$ as a function on the charts $\partial\Delta_{\lambda}^{\vee}\cap U_{w}^{\infty}$ with $\langle w,m\rangle=1$ . We say $u$ is a locally convex function if all $u_{m}$ are convex on their corresponding charts.

00QY

Remark 3.23. One can reconstruct $u$ from the local functions $\{u_{m}\}$ as long as their mutural differences define a correct cocycle $\{m-m^{\prime}\}$ . Thus this definition has the intrinsic local feature we desire, in analogy with the notion of Kähler potentials.

00QZ

Remark 3.24. For fixed $w$ and $m,m^{\prime}$ satisfying $\langle m,w\rangle=\langle m^{\prime},w\rangle=1$ , the convexity of $u_{m}$ and $u_{m^{\prime}}$ on the $w$ -chart are equivalent because $m-m^{\prime}$ is an affine function. However, on the overlap of the $w$ -chart and the $w^{\prime}$ -chart, if $u_{m}$ is convex in one chart it is not automatically convex in the other.

00R0

Remark 3.25. If a convex function is not sufficiently regular, there can be a null set of points at which the subgradient is not unique. Later we will abuse language to use the word gradient to refer to any choice of subgradient.

00R1

Proposition 3.26. If $u$ satisfies the extension property, then $u$ is locally convex.

00R2

Proof. Let $\langle m,w\rangle=1$ , and consider the function $u_{m}$ on the chart $\partial\Delta_{\lambda}^{\vee}\cap U_{w}^{\infty}$ . Given $x$ in the chart, we need to find $\vec{p}$ such that

u_{m}(y)-u_{m}(x)\geq\vec{p}\cdot(y-x)_{w},

where $\vec{p}$ is a covector, and $(y-x)_{w}$ refers to the representation of $y-x$ in the local coordinates $x^{m_{1}},\ldots x^{m_{n}}$ ; after identifying $x^{m_{1}},\ldots x^{m_{n}}$ as coordinates on the plane $m^{\perp}=\{\langle m,x^{\prime}\rangle=0\}$ , we may regard $(y-x)_{w}$ as an element of $m^{\perp}$ , and according to the decomposition $N_{\mathbb{R}}=m^{\perp}\oplus\mathbb{R}w$ ,

y-x=(y-x)_{w}+\langle y-x,m\rangle w.

Since the convexity of $u_{m}$ and $u_{m^{\prime}}$ are equivalent in the $w$ -chart if $\langle m,w\rangle=\langle m,w\rangle=1$ , we may assume $L_{\lambda}(x)$ is attained by $\langle m,x\rangle+\lambda(m)$ . By the extension property and Prop. 3.19, there is some $p\in\Delta$ , such that

u(y)-u(x)\geq\langle p,y-x\rangle,

hence

u_{m}(y)-u_{m}(x)\geq\langle p-m,y-x\rangle=\langle p-m,(y-x)_{w}\rangle+\langle y-x,m\rangle\langle p-m,w\rangle.

Since $p\in\Delta$ , we have $\langle p,w\rangle\leq 1=\langle m,w\rangle.$ Since $L_{\lambda}(x)$ is attained by $\langle m,x\rangle+\lambda(m)$ , and the polytope $\Delta_{\lambda}^{\vee}$ lies in the half space $\{\langle m,\rangle+\lambda(m)\leq 0\}$ , we have

\langle m,y\rangle+\lambda(m)\leq 0=\langle m,x\rangle+\lambda(m).

Combining the above

u_{m}(x)-u_{m}(y)\geq\langle p-m,(y-x)_{w}\rangle+\langle y-x,m\rangle\langle p-m,w\rangle\geq\langle p-m,(y-x)_{w}\rangle,

so we have produced $\vec{p}$ as required. ∎