$D_5^{(1)}$- Geometric Crystal corresponding to the Dynkin spin node $i=5$ and its ultra-discretization

TL;DR

This paper constructs a positive geometric crystal for the affine Lie algebra $D_5^{(1)}$, explicitly describes its ultra-discretization, and proves it matches the limit of a coherent family of perfect crystals, confirming a conjecture in this case.

Contribution

It provides the first explicit construction of a positive geometric crystal for $D_5^{(1)}$ and establishes its ultra-discretization as the limit of perfect crystals, confirming the conjecture for this algebra.

Findings

Constructed a positive geometric crystal $V(D_5^{(1)})$ for $D_5^{(1)}$.

Defined explicit $0$-action on the perfect crystal $B^{5, l}$.

Proved the ultra-discretization of $V(D_5^{(1)})$ is isomorphic to $B^{5, \, ext{infinity}}$.

Abstract

Let $g$ be an affine Lie algebra with index set $I = \{0, 1, 2, \cdots , n\}$ and $g^L$ be its Langlands dual. It is conjectured that for each Dynkin node $i \in I \setminus \{0\}$ the affine Lie algebra $g$ has a positive geometric crystal whose ultra-discretization is isomorphic to the limit of certain coherent family of perfect crystals for $g^L$. In this paper we construct a positive geometric crystal $V(D_5^{(1)})$ in the level zero fundamental spin $D_5^{(1)}$- module $W(\varpi_5)$. Then we define explicit $0$-action on the level $l$ known $D_5^{(1)}$- perfect crystal $B^{5, l}$ and show that $\{B^{5, l}\}_{l \geq 1}$ is a coherent family of perfect crystals with limit $B^{5, \infty}$. Finally we show that the ultra-discretization of $V(D_5^{(1)})$ is isomorphic to $B^{5, \infty}$ as crystals which proves the conjecture in this case.

Equations

$$\phi_{i}\left(\left(\begin{array}[]{cc}c&0\\ 0&c^{-1}\end{array}\right)\right)=c^{\check{\alpha}_{i}},\,\phi_{i}\left(\left(\begin{array}[]{cc}1&t\\ 0&1\end{array}\right)\right)=\exp(te_{i}),\,\phi_{i}\left(\left(\begin{array}[]{cc}1&0\\ t&1\end{array}\right)\right)=\exp(tf_{i}).$$

$$B_{\mathbf{i}}^{-}:=\{Y_{\mathbf{i}}(c_1,c_2,\cdots ,c_l):=Y_{i_1}(c_1)Y_{i_2}(c_2)\cdots Y_{i_l}(c_l)|c_1,c_2,\cdots ,c_l\in {\mathbb{C}}^{\times }\}\subset B^{-},$$

$$e_k^c(Y_{\mathbf{i}}(c_1,\cdots ,c_l))=Y_{\mathbf{i}}({\mathcal{C}}_1,\cdots ,{\mathcal{C}}_l)),$$

$${\mathcal{C}}_j:=c_j\cdot \frac{\sum _{1\le m\le j,i_m=k}\frac{c}{c_1^{a_{i_1,k}}\cdots c_{m-1}^{a_{i_{m-1},k}}{c}_m}+\sum _{j<m\le k,i_m=k}\frac{1}{c_1^{a_{i_1,k}}\cdots c_{m-1}^{a_{i_{m-1},k}}{c}_m}}{\sum _{1\le m<j,i_m=k}\frac{c}{c_1^{a_{i_1,k}}\cdots c_{m-1}^{a_{i_{m-1},k}}{c}_m}+{\sum }_{j\le m\le k,i_m=k}\frac{1}{c_1^{a_{i_1,k}}\cdots c_{m-1}^{a_{i_{m-1},k}}{c}_m}},$$

$${\epsilon }_k(Y_{\mathbf{i}}(c_1,\cdots ,c_l))=\sum _{1\le m\le l,i_m=k}\frac{1}{c_1^{a_{i_1,k}}\cdots c_{m-1}^{a_{i_{m-1},k}}{c}_m},$$

$${\gamma }_k(Y_{\mathbf{i}}(c_1,\cdots ,c_l))=c_1^{a_{i_1,k}}\cdots c_l^{a_{i_l,k}},$$

$$x\times y⟼x+y,\frac{x}{y}⟼x-y,x+y⟼\max \{x,y\}.$$

$$\sigma :0\mapsto 5,1\mapsto 4,2\mapsto 3,3\mapsto 2,4\mapsto 0,5\mapsto 1.$$

$$\{v=(i_1,i_2,i_3,i_4,i_5)|i_j\in \{+,-\},i_1{i}_2{i}_3{i}_4{i}_5=+\}.$$

$$\begin{split}&e_{k}(i_{1},i_{2},i_{3},i_{4},i_{5})=\left\{\begin{array}[]{lllll}(-,-,i_{3},i_{4},i_{5})&&\text{ if }k=0,\ (i_{1},i_{2})=(+,+)\\ (i_{1},\ldots,+,-,\ldots,i_{5})&&\text{ if }k\neq 0,\ k\neq 5,\ (i_{k},i_{k+1})=(-,+)\\ \hskip 34.14322ptk\hskip 7.96674ptk+1\\ (i_{1},i_{2},i_{3},+,+)&&\text{ if }k=5,\ (i_{4},i_{5})=(-,-)\\ 0&&\text{ otherwise.}\end{array}\right.\\ &f_{k}(i_{1},i_{2},i_{3},i_{4},i_{5})=\left\{\begin{array}[]{lllll}(+,+,i_{3},i_{4},i_{5})&&\text{ if }k=0,\ (i_{1},i_{2})=(-,-)\\ (i_{1},\ldots,-,+,\ldots,i_{5})&&\text{ if }k\neq 0,\ k\neq 5,\ (i_{k},i_{k+1})=(+,-)\\ \hskip 34.14322ptk\hskip 7.96674ptk+1\\ (i_{1},i_{2},i_{3},-,-)&&\text{ if }k=5,\ (i_{4},i_{5})=(+,+)\\ 0&&\text{ otherwise.}\end{array}\right.\end{split}$$

$$\begin{split}&\langle{\check{\alpha}_{k}},{\rm wt}(v)\rangle=\left\{\begin{array}[]{lllll}1&&\text{ if }\ k=0,\ (i_{1},i_{2})=(-,-)\\ &&\text{ or }k\neq 0,\ k\neq 5,\ (i_{k},i_{k+1})=(+,-)\\ &&\text{ or }k=5,(i_{4},i_{5})=(+,+)\\ -1&&\text{ if }\ k=0,\ (i_{1},i_{2})=(+,+)\\ &&\text{ or }k\neq 0,\ k\neq 5,(i_{k},i_{k+1})=(-,+)\\ &&\text{ or }k=5,\ (i_{4},i_{5})=(-,-)\\ 0&&\text{ otherwise.}\end{array}\right.\\ \end{split}$$

$$t({\varpi }_5)=\sigma s_4{s}_3{s}_2{s}_5{s}_3{s}_4{s}_1{s}_2{s}_3{s}_5=\sigma w_1,$$

$$t(\check{{\varpi }_5})=\sigma s_5{s}_3{s}_2{s}_4{s}_3{s}_5{s}_0{s}_2{s}_3{s}_4=\sigma w_2,$$

$${\mathcal{V}}_1$$

$$\displaystyle\hskip 28.45274ptY_{1}(x_{1}^{(1)})Y_{2}(x_{2}^{(1)})Y_{3}(x_{3}^{(1)})Y_{5}(x_{5}^{(1)})(+,+,+,+,+)|x_{m}^{(l)}\in\mathbb{C}^{\times}\big{\}},$$

$${\mathcal{V}}_2$$

$$\displaystyle\hskip 28.45274ptY_{0}(y_{0}^{(1)})Y_{2}(y_{2}^{(1)})Y_{3}(y_{3}^{(1)})Y_{4}(y_{4}^{(1)})(-,+,+,+,-)|y_{m}^{(l)}\in\mathbb{C}^{\times}\big{\}},$$

$$Y_k(c)=(1+\frac{f_k}{c}){\check{\alpha }}_k(c)=(1+\frac{f_k}{c})c^{{\check{\alpha }}_k}\text{for all}k\in I.$$

$$V_2(y)=a(x)\sigma (V_1(x)).$$

$$a(x)$$

$$y_5^{(2)}$$

$$y_4^{(2)}$$

$$y_3^{(3)}$$

$$y_3^{(2)}$$

$$y_3^{(1)}$$

$$y_2^{(2)}$$

$$\bar{\sigma }:{\mathcal{V}}_1\to {\mathcal{V}}_2,$$

$$V_1(x)\mapsto V_2(y).$$

$${\bar{\sigma }}^{-1}:{\mathcal{V}}_2\to {\mathcal{V}}_1,$$

$$V_2(y)\mapsto V_1(x)$$

$$x_5^{(2)}$$

$$x_4^{(2)}$$

$$x_3^{(3)}$$

$$x_3^{(2)}$$

$$x_3^{(1)}$$

$$x_2^{(2)}$$

$$e_k^c(V_1(x))$$

Full text

KCM is partially supported by the Simons Foundation Grant #307555.

1. Introduction

Let $\mathfrak{g}$ be an affine Lie algebra [11] with Cartan datum $\{A,\Pi,\check{\Pi},P,\check{P}\}$ and index set $I=\{0,1,\cdots,n\}$ where $A=(a_{ij})_{i,j\in I}$ is the affine GCM, $\Pi=\{\alpha_{i}\mid i\in I\}$ is the set of simple roots, $\check{\Pi}=\{\check{\alpha_{i}}\mid i\in I\}$ is the set of simple coroots, $P$ and $\check{P}$ are the weight lattice, and coweight lattice respectively. Let $\mathfrak{t}=\mathbb{C}\otimes_{\mathbb{Z}}\check{P}$, $\bf c$, $\delta$, and $\{\Lambda_{i}\mid i\in I\}$ denote the Cartan subalgebra, the canonical central element, the null root and the set of fundamental weights respectively. Note that $\alpha_{j}(\check{\alpha}_{i})=a_{ij}$ and $\Lambda_{j}(\check{\alpha_{i}})=\delta_{ij}$ and $\mathfrak{t}=\text{span}_{\mathbb{C}}\{\check{\alpha}_{i},d\mid i\in I\}$ where $d$ is a degree derivation. Then $P=\oplus_{j\in I}\mathbb{Z}\Lambda_{j}\oplus\mathbb{Z}\delta\subset\mathfrak{t}^{*}$, $\check{P}=\oplus_{i\in I}\mathbb{Z}\check{\alpha}_{i}\oplus\mathbb{Z}d\subset\mathfrak{t}$ and $P_{cl}=P/\mathbb{Z}\delta$ is called the classical weight lattice. The set $P^{+}=\{\lambda\in P\mid\lambda(\check{\alpha}_{i})\in\mathbb{Z}_{\geq 0}\;\text{for}\;\text{all}\;i\in I\}$ (resp. $P_{cl}^{+}=\{\lambda\in P_{cl}\mid\lambda(\check{\alpha}_{i})\in\mathbb{Z}_{\geq 0}\;\text{for}\;\text{all}\;i\in I\}$) is called the set of (affine) dominant (resp. classical dominant) weights and we say that $\lambda\in P^{+}$ or $P_{cl}^{+}$ has level $l=\lambda({\bf c})$. We denote $(P^{+})_{l}$ (resp. $(P_{cl}^{+})_{l}$) to be the set of affine (resp. classical) dominant weights of level $l$. We denote $\mathfrak{t}_{cl}^{*}=\mathfrak{t}^{*}/{{\mathbb{C}}\delta}$ and $(\mathfrak{t}_{cl}^{*})_{0}=\{\lambda\in\mathfrak{t}_{cl}^{*}\mid\langle{\bf c},\lambda\rangle=0\}$. We also denote $\mathfrak{g}_{i}$ to be the subalgebra of $\mathfrak{g}$ with index set $I_{i}=I\setminus\{i\}$ which is a finite dimensional semisimple Lie algebra. The Weyl group $W$ of $\mathfrak{g}$ is generated by the simple reflections $\{s_{i}\mid i\in I\}$. The sets $\Delta$, $\Delta_{+}$ , $\Delta^{re}:=\{w(\alpha_{i})|w\in W,\,\,i\in I\}$ and $\Delta^{re}_{+}=\Delta_{+}\cap\Delta^{re}$ are called the set of roots, postive roots, real roots and positive real roots respectively. In this paper we will assume $\mathfrak{g}$ to be simply laced which implies that the affine GCM $A=(a_{ij})_{i,j\in I}$ is symmetric.

Let $\mathfrak{g}^{\prime}$ be the derived Lie algebra of $\mathfrak{g}$ and let $G$ be the Kac-Moody group associated with $\mathfrak{g}^{\prime}$([12, 19]). Let $U_{\alpha}:=\exp\mathfrak{g}_{\alpha}$ $(\alpha\in\Delta^{re})$ be the one-parameter subgroup of $G$. The group $G$ is generated by $U_{\alpha}$ $(\alpha\in\Delta^{re})$. Let $U^{\pm}:=\langle U_{\pm\alpha}|\alpha\in\Delta^{re}_{+}\rangle$ be the subgroup generated by $U_{\pm\alpha}$ ($\alpha\in\Delta^{re}_{+}$). For any $i\in I$, there exists a unique homomorphism; $\phi_{i}:SL_{2}(\mathbb{C})\rightarrow G$ such that

[TABLE]

where $c\in\mathbb{C}^{\times}$ and $t\in\mathbb{C}$. Set $\check{\alpha}_{i}(c):=c^{\check{\alpha}_{i}}$, $x_{i}(t):=\exp{(te_{i})}$, $y_{i}(t):=\exp{(tf_{i})}$, $G_{i}:=\phi_{i}(SL_{2}(\mathbb{C}))$, $H_{i}:=\phi_{i}(\{{\rm diag}(c,c^{-1})\mid c\in\mathbb{C}\setminus\{0\}\})$. Let $H$ be the subgroup of $G$ generated by $H_{i}$’s with the Lie algebra $\mathfrak{t}$. Then $H$ is called a maximal torus in $G$, and $B^{\pm}=U^{\pm}H$ are the Borel subgroups of $G$. The element $\bar{s}_{i}:=x_{i}(-1)y_{i}(1)x_{i}(-1)\in N_{G}(H)$ is a representative of $s_{i}\in W=N_{G}(H)/H$.

The geometric crystal for the simply laced affine Lie algebra $\mathfrak{g}$ is defined as follows.

Definition 1.1**.**

([1],[16]) The geometric crystal for the simply laced affine Lie algebra $\mathfrak{g}$ is a quadruple $\mathcal{V}(\mathfrak{g})=(X,\{e_{i}\}_{i\in I},\{\gamma_{i}\}_{i\in I},$ $\{\varepsilon_{i}\}_{i\in I})$, where $X$ is an ind-variety, $e_{i}:\mathbb{C}^{\times}\times X\longrightarrow X$ $((c,x)\mapsto e^{c}_{i}(x))$ are rational $\mathbb{C}^{\times}$-actions and $\gamma_{i},\varepsilon_{i}:X\longrightarrow\mathbb{C}$ $(i\in I)$ are rational functions satisfying the following:

1. (1)

$\{1\}\times X\subset{\rm dom}(e_{i})\;{\rm for}\;{\rm any}\;i\in I.$ 2. (2)

$\gamma_{j}(e^{c}_{i}(x))=c^{a_{ij}}\gamma_{j}(x).$ 3. (3)

\{e_{i}\}_{i\in I}\;{\rm satisfy\;the\;following\;relations}:\\ \begin{array}[]{lll}&\quad e^{c_{1}}_{i}e^{c_{2}}_{j}=e^{c_{2}}_{j}e^{c_{1}}_{i}&{\rm if}\,\,a_{ij}=a_{ji}=0,\\ &\quad e^{c_{1}}_{i}e^{c_{1}c_{2}}_{j}e^{c_{2}}_{i}=e^{c_{2}}_{j}e^{c_{1}c_{2}}_{i}e^{c_{1}}_{j}&{\rm if}\,\,a_{ij}=a_{ji}=-1,\\ \end{array} 4. (4)

$\varepsilon_{i}(e_{i}^{c}(x))=c^{-1}\varepsilon_{i}(x)$ and $\varepsilon_{i}(e_{j}^{c}(x))=\varepsilon_{i}(x)\qquad{\rm if}\,a_{i,j}=a_{j,i}=0.$

For fixed $i\in I$, let $G^{i}$ be the reductive algebraic group with Lie algebra $\mathfrak{g}_{i}$ and $B^{i}$, $W^{i}$ be its Borel subgroup, Weyl group respectively. We consider the flag variety $X^{i}:=G^{i}/{B^{i}}$. For $w\in W^{i}$, the Schubert cell $X^{i}_{w}$ associated with $w$ has a natural $\mathfrak{g}_{i}$-geometric crystal structure [1, 16]. Let $w=s_{i_{1}}s_{i_{2}}\cdots s_{i_{l}}$ be a reduced expression. For ${\bf i}:=(i_{1},i_{2},\cdots,i_{l})$, set

[TABLE]

where $Y_{j}(c):=y_{j}(\frac{1}{c}){\check{\alpha}}_{j}(c)=y_{j}(\frac{1}{c})c^{{\check{\alpha}}_{j}}$. Then we have the following result.

Theorem 1.2**.**

[1, 16]** The set $B_{\bf i}^{-}$ with the explicit actions of   $e^{c}_{k}$, $\varepsilon_{k}$, and $\gamma_{k}$ , for $k\in I_{i},c\in\mathbb{C}^{\times}$ given by:

[TABLE]

is a geometric crystal isomorphic to $X^{i}_{w}$.

The geometric crystal $\mathcal{V}(\mathfrak{g})=(X,\{e_{i}\}_{i\in I},\{\gamma_{i}\}_{i\in I},\{\varepsilon_{i}\}_{i\in I})$ is said to be positive if it has a positive structure [1, 16, 10]. Roughly speaking this means that each of the rational maps $e^{c}_{i}$, $\varepsilon_{i}$ and $\gamma_{i}$ are given by ratio of polynomial functions with positive coefficients. For example, $B_{\bf i}^{-}$ is a positive geometric crystal.

For a dominant weight $\lambda\in P^{+}$ of level $l$, Kashiwara defined the crystal base $(L(\lambda),B(\lambda))$ [6] (also see [15]) for the integrable highest weight $\mathfrak{g}$-module $V(\lambda)$. As shown in [3], the crystal $B(\lambda)$ can be realized as a set of paths in the semi-infinite tensor product $\cdots\otimes B^{l}\otimes B^{l}\otimes B^{l}$ where $B^{l}$ is a perfect crystal of level $l$. This is called the path realization of the crystal $B(\lambda)$. When a family of perfect crystals $\{B^{l}\}_{l\geq 1}$ are coherent [5], it has a limit $B^{\infty}$. The positive geometric crystals are related to Kashiwara crystals via the ultra-discretization functor $\mathcal{UD}$ [1, 16] which transforms positive rational functions to piecewise-linear functions by the simple correspondence:

[TABLE]

It was conjectured in [10] that for each affine Lie algebra $\mathfrak{g}$ and each Dynkin index $k\in I\setminus\{0\}$, there exists a positive geometric crystal $\mathcal{V}(\mathfrak{g})=(X,\{e_{i}\}_{i\in I},\{\gamma_{i}\}_{i\in I},\\ \{\varepsilon_{i}\}_{i\in I})$ whose ultra-discretization $\mathcal{UD}(\mathcal{V})$ is isomorphic to the limit $B^{\infty}$ of a coherent family of perfect crystals for the Langlands dual $\mathfrak{g}^{L}$. If $\mathfrak{g}$ is simply laced then the Langland dual is $\mathfrak{g}$ itself. So far this conjecture has been proved for the Dynkin index $k=1$ and $\mathfrak{g}=A_{n}^{(1)},B_{n}^{(1)},C_{n}^{(1)},D_{n}^{(1)},A_{2n-1}^{(2)},A_{2n}^{(2)},$ $D_{n+1}^{(2)}$ [10], $\mathfrak{g}=D_{4}^{(3)}$ [2], $\mathfrak{g}=G_{2}^{(1)}$ [17]. For $k>1$ , this conjecture has been shown to hold only for $\mathfrak{g}=A_{n}^{(1)}$ [13, 14].

In this paper we prove the conjecture in [10] for $\mathfrak{g}=D_{5}^{(1)}$ and Dynkin idex $k=5$, the spin node. In Section 2, we construct a positive geometric crystal $\mathcal{V}(D_{5}^{(1)})$ in the level zero fundamental spin module $W(\varpi_{5})$. In Section 3, for $l\geq 1$ we coordinatize the perfect crystal $B^{5,l}$ for $D_{5}^{(1)}$ given in [4] and define an explicit [math]-action. Then we show that the family of perfect crystals $\{B^{5,l}\}_{l\geq 1}$ is a coherent family and determine its limit $B^{5,\infty}$. In the last section we ultra-discretize the positive geometric crystal $\mathcal{V}(D_{5}^{(1)})$ and show that it is isomorphic to $B^{5,\infty}$ as crystals.

2. Affine Geometric Crystal $\mathcal{V}(D_{5}^{(1)})$

From now on we assume $\mathfrak{g}$ to be the affine Lie algebra $D_{5}^{(1)}$ with index set $I=\{0,1,2,3,4,5\}$, Cartan matrix $A=(a_{ij})_{i,j\in I}$ where $a_{ii}=2,a_{j,j+1}=-1=a_{j+1,j},\;j=1,2,3,a_{02}=a_{20}=a_{35}=a_{53}=-1,a_{ij}=0$ otherwise and Dynkin diagram:

012345

Let $\{\alpha_{0},\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\},\ \{\check{\alpha_{0}},\check{\alpha_{1}},\check{\alpha_{2}},\check{\alpha_{3}},\check{\alpha_{4}},\check{\alpha_{5}}\}$ and $\{\Lambda_{0},\Lambda_{1},\Lambda_{2},\Lambda_{3},\Lambda_{4},\Lambda_{5}\}$ denote the set of simple roots, simple coroots and fundamental weights, respectively. Then ${\bf c}=\check{\alpha_{0}}+\check{\alpha_{1}}+2\check{\alpha_{2}}+2\check{\alpha_{3}}+\check{\alpha_{4}}+\check{\alpha_{5}}$ and $\delta=\alpha_{0}+\alpha_{1}+2\alpha_{2}+2\alpha_{3}+\alpha_{4}+\alpha_{5}$ are the central element and null root respectively. The sets $P_{cl}=\oplus_{j=0}^{5}\mathbb{Z}\Lambda_{j}$ and $P=P_{cl}\oplus\mathbb{Z}\delta$ are called classical weight lattice and weight lattice respectively.

We consider the Dynkin diagram automorphism111We thank T. Nakashima for suggesting it to us. $\sigma$ defined by

[TABLE]

012345$\sigma$012354D_{5}$$D_{5}

Let $\mathfrak{g}_{j}$ (resp. $\sigma(\mathfrak{g})_{j})$) be the subalgebra of $\mathfrak{g}$ (resp. $\sigma(\mathfrak{g})$) with index set $I_{j}=I\setminus\{j\}$. Then observe that $\mathfrak{g}_{0}$ as well as $\mathfrak{g}_{1}$ and $\sigma(\mathfrak{g})_{1}$ are isomorphic to $D_{5}$.

Let $W(\varpi_{5})$ be the level [math] fundamental $U^{\prime}_{q}(\mathfrak{g})$-module associated with the level [math] weight $\varpi_{5}=\Lambda_{5}-\Lambda_{0}$ [8]. By [[8], Theorem 5.17], $W(\varpi_{5})$ is a finite-dimensional irreducible integrable $U^{\prime}_{q}(\mathfrak{g})$-module and has a global basis with a simple crystal. Thus, we can consider the specialization $q=1$ and obtain the finite-dimensional $D_{5}^{(1)}$-module $W(\varpi_{5})$, which we call the fundamental $D_{5}^{(1)}$- module and use the same notation as above. Below we give the explicit description of $W(\varpi_{5})$.

2.1. Fundamental Representation $W(\varpi_{5})$** for $D_{5}^{(1)}$**

The fundamental $D_{5}^{(1)}$-module $W(\varpi_{5})$ is a 16-dimensional module with the basis

[TABLE]

The actions of the generators $e_{k}$, and $f_{k}$, $0\leq k\leq 5$ of $D_{5}^{(1)}$, on the basis vectors are given as follows.

[TABLE]

Furthermore, we observe that

[TABLE]

Note that in $W(\varpi_{5})$, we have $(+,+,+,+,+)$ (resp. $(-,+,+,+,-)$) is a $\mathfrak{g}_{0}$ (resp. $\mathfrak{g}_{1}$) highest weight vector with weight $\varpi_{5}=\Lambda_{5}-\Lambda_{0}$ (resp. $\check{\varpi_{5}}:=\Lambda_{4}-\Lambda_{1}$). We define $\sigma(\Lambda_{j})=\Lambda_{\sigma(j)}$ for $j\in I$. Then we define the action of $\sigma$ on $W(\varpi_{5})$ by $\sigma(v)=v^{\prime}$ if $\sigma({\rm wt}(v))={\rm wt}(v^{\prime})$.

2.2. Affine Geometric Crystal $\mathcal{V}(D_{5}^{(1)})$** in $W(\varpi_{5})$**

Now we will construct the affine geometric crystal $\mathcal{V}(D_{5}^{(1)})$ in $W(\varpi_{5})$ explicitly. For $\xi\in(\mathfrak{t}^{*}_{\text{cl}})_{0}$, let $t(\xi)$ be the translation as in [[8], Sect 4]. Define simple reflections $s_{k}(\lambda):=\lambda-\lambda({\check{\alpha}_{k}})\alpha_{k},k\in I$ and let $W=\langle s_{k}\mid k\in I\rangle$ be the Weyl group for $D_{5}^{(1)}$. Then we have

[TABLE]

where $w_{1}=s_{4}s_{3}s_{2}s_{5}s_{3}s_{4}s_{1}s_{2}s_{3}s_{5}\in W^{0}$ and $w_{2}=s_{5}s_{3}s_{2}s_{4}s_{3}s_{5}s_{0}s_{2}s_{3}s_{4}\in W^{1}$. Associated with these Weyl group elements $w_{1},w_{2}\in W$, we define algebraic varieties $\mathcal{V}_{1},\mathcal{V}_{2}\subset W(\varpi_{5})$ as follows.

[TABLE]

where $x=(x_{4}^{(2)},x_{3}^{(3)},x_{2}^{(2)},x_{5}^{(2)},x_{3}^{(2)},x_{4}^{(1)},x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},x_{5}^{(1)})$ and $y=(y_{5}^{(2)},y_{3}^{(3)},y_{2}^{(2)},\\ y_{4}^{(2)},y_{3}^{(2)},y_{5}^{(1)},y_{0}^{(1)},y_{2}^{(1)},y_{3}^{(1)},y_{4}^{(1)})$.

From the explicit actions of $f_{k}$’s on $W(\varpi_{5})$, we observe that $f_{k}^{2}=0$, for all $k\in I$. Therefore, we have

[TABLE]

Thus we have the explicit forms of $V_{1}(x)$ and $V_{2}(y)$ as follows.

V_{1}(x)=x_{5}^{(2)}x_{5}^{(1)}(+,+,+,+,+)+\big{(}x_{3}^{(3)}x_{5}^{(1)}+\frac{x_{3}^{(3)}x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}}\big{)}(+,+,+,-,-)+\big{(}x_{4}^{(2)}x_{5}^{(1)}+\frac{x_{4}^{(2)}x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}}+\frac{x_{4}^{(2)}x_{2}^{(2)}x_{3}^{(1)}}{x_{3}^{(3)}}+\frac{x_{4}^{(2)}x_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}}\big{)}(+,+,-,+,-)+\big{(}x_{5}^{(1)}+\frac{x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}}+\frac{x_{2}^{(2)}x_{3}^{(1)}}{x_{3}^{(3)}}+\frac{x_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}}+\frac{x_{2}^{(2)}x_{2}^{(1)}}{x_{4}^{(2)}}\big{)}(+,+,-,-,+)+\big{(}x_{4}^{(2)}x_{3}^{(1)}+\frac{x_{4}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(2)}}+\frac{x_{4}^{(2)}x_{4}^{(1)}x_{1}^{(1)}}{x_{2}^{(2)}}\big{)}(+,-,+,+,-)+\big{(}x_{3}^{(1)}+\frac{x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(2)}}+\frac{x_{4}^{(1)}x_{1}^{(1)}}{x_{2}^{(2)}}+\frac{x_{3}^{(3)}x_{2}^{(1)}}{x_{4}^{(2)}}+\frac{x_{3}^{(3)}x_{3}^{(2)}x_{1}^{(1)}}{x_{4}^{(2)}x_{2}^{(2)}}\big{)}(+,-,+,-,+)+\big{(}x_{2}^{(1)}+\frac{x_{3}^{(2)}x_{1}^{(1)}}{x_{2}^{(2)}}+\frac{x_{5}^{(2)}x_{1}^{(1)}}{x_{3}^{(3)}}\big{)}(+,-,-,+,+)+x_{1}^{(1)}(+,-,-,-,-)+x_{4}^{(2)}x_{4}^{(1)}(-,+,+,+,-)+\\ \big{(}x_{4}^{(1)}+\frac{x_{3}^{(3)}x_{3}^{(2)}}{x_{4}^{(2)}}\big{)}(-,+,+,-,+)+\big{(}x_{3}^{(2)}+\frac{x_{2}^{(2)}x_{5}^{(2)}}{x_{3}^{(3)}}\big{)}(-,+,-,+,+)+x_{5}^{(2)}(-,-,+,+,+)+x_{2}^{(2)}(-,+,-,-,-)+x_{3}^{(3)}(-,-,+,-,-)+x_{4}^{(2)}(-,-,-,+,-)+(-,-,-,-,+),

V_{2}(y)=y_{4}^{(2)}y_{4}^{(1)}(-,+,+,+,-)+\big{(}y_{3}^{(3)}y_{4}^{(1)}+\frac{y_{3}^{(3)}y_{3}^{(2)}y_{3}^{(1)}}{y_{4}^{(2)}}\big{)}(-,+,+,-,+)+\big{(}y_{5}^{(2)}y_{4}^{(1)}+\frac{y_{5}^{(2)}y_{3}^{(2)}y_{3}^{(1)}}{y_{4}^{(2)}}+\frac{y_{5}^{(2)}y_{2}^{(2)}y_{3}^{(1)}}{y_{3}^{(3)}}+\frac{y_{5}^{(2)}y_{2}^{(2)}y_{5}^{(1)}y_{2}^{(1)}}{y_{3}^{(3)}y_{3}^{(2)}}\big{)}(-,+,-,+,+)+\big{(}y_{4}^{(1)}+\frac{y_{3}^{(2)}y_{3}^{(1)}}{y_{4}^{(2)}}+\frac{y_{2}^{(2)}y_{3}^{(1)}}{y_{3}^{(3)}}+\frac{y_{2}^{(2)}y_{5}^{(1)}y_{2}^{(1)}}{y_{3}^{(3)}y_{3}^{(2)}}+\frac{y_{2}^{(2)}y_{2}^{(1)}}{y_{5}^{(2)}}\big{)}(-,+,-,-,-)+\big{(}y_{5}^{(2)}y_{3}^{(1)}+\frac{y_{5}^{(2)}y_{5}^{(1)}y_{2}^{(1)}}{y_{3}^{(2)}}+\frac{y_{5}^{(2)}y_{5}^{(1)}y_{0}^{(1)}}{y_{2}^{(2)}}\big{)}(-,-,+,+,+)+\big{(}y_{3}^{(1)}+\frac{y_{5}^{(1)}y_{2}^{(1)}}{y_{3}^{(2)}}+\frac{y_{5}^{(1)}y_{0}^{(1)}}{y_{2}^{(2)}}+\frac{y_{3}^{(3)}y_{2}^{(1)}}{y_{5}^{(2)}}+\frac{y_{3}^{(3)}y_{3}^{(2)}y_{0}^{(1)}}{y_{5}^{(2)}y_{2}^{(2)}}\big{)}(-,-,+,-,-)+\big{(}y_{2}^{(1)}+\frac{y_{3}^{(2)}y_{0}^{(1)}}{y_{2}^{(2)}}+\frac{y_{4}^{(2)}y_{0}^{(1)}}{y_{3}^{(3)}}\big{)}(-,-,-,+,-)+y_{0}^{(1)}(-,-,-,-,+)+y_{5}^{(2)}y_{5}^{(1)}(+,+,+,+,+)+\\ \big{(}y_{5}^{(1)}+\frac{y_{3}^{(3)}y_{3}^{(2)}}{y_{5}^{(2)}}\big{)}(+,+,+,-,-)+\big{(}y_{3}^{(2)}+\frac{y_{2}^{(2)}y_{4}^{(2)}}{y_{3}^{(3)}}\big{)}(+,+,-,+,-)+y_{4}^{(2)}(+,-,+,+,-)+y_{2}^{(2)}(+,+,-,-,+)+y_{3}^{(3)}(+,-,+,-,+)+y_{5}^{(2)}(+,-,-,+,+)+(+,-,-,-,-).

Now for a given $x$ we solve the equation

[TABLE]

where $a(x)$ is a rational function in $x$ and the action of $\sigma$ on $V_{1}(x)$ is induced by its action on $W(\varpi_{5})$. Though this equation is over-determined, it can be solved uniquely by comparing the coefficients of the basis vectors of $W(\varpi_{5})$. We give the explicit solutions of $a(x)$, and the variables $y_{m}^{(l)}$ below.

Lemma 2.1**.**

The rational function $a(x)$ and the complete solution of (2.1) is:

[TABLE]

[TABLE]

[TABLE]

Using Lemma 2.1 we define the map

[TABLE]

[TABLE]

Now we have the following result.

Proposition 2.2**.**

The map $\bar{\sigma}:\mathcal{V}_{1}\rightarrow\mathcal{V}_{2}$ is a bi-positive birational isomorphism with the inverse positive rational map

[TABLE]

[TABLE]

given by

[TABLE]

[TABLE]

[TABLE]

Proof.

The fact that $\bar{\sigma}$ is a bi-positive birational map follows from the explicit formulas. The rest follows by direct calculations. ∎

It is known that $\mathcal{V}_{1}$ (resp. $\mathcal{V}_{2}$) has the structure of a $\mathfrak{g}_{0}$ (resp. $\mathfrak{g}_{1}$) positive geometric crystal ([1], [16], [10]). Indeed, note that taking the sesquence ${\bf i}=(4,3,2,5,3,4,1,2,3,5)$ the explicit actions of  $e_{k}^{c},\ \gamma_{k},\ \varepsilon_{k}$ on $V_{1}(x)$ for $k=1,2,3,4,5$ are given by Theorem 1.2 as follows.

[TABLE]

By choosing ${\bf i}=(5,3,2,4,3,5,0,2,3,4)$, we also have the following explicit actions of  $\bar{e_{k}}^{c},\ \bar{\gamma}_{k},\ \bar{\varepsilon}_{k},\ k=0,2,3,4,5$ on $V_{2}(y)$ by Theorem 1.2.

[TABLE]

[TABLE]

Proposition 2.3**.**

*The following relations hold.

(i) $\bar{\sigma}e_{2}^{c}=\bar{e_{3}}^{c}\bar{\sigma}$ ,

(ii) $\bar{\sigma}e_{3}^{c}=\bar{e_{2}}^{c}\bar{\sigma}$.*

Proof.

We will prove the relation $(i)$ and relation $(ii)$ can be shown similarly. Set $e_{2}^{c}(V_{1}(x))=V_{1}(z)$, $\bar{\sigma}(V_{1}(z))=V_{2}(y^{\prime})$, $\bar{\sigma}(V_{1}(x))=V_{2}(y)$ and $\bar{e_{3}}^{c}(V_{2}(y))=V_{2}(w)$. We need to show that $y_{m}^{(l)^{\prime}}=w_{m}^{(l)}$ for $(l,m)\in\{(1,0),(1,2),(1,3),(1,4),(1,5),(2,2),\\ (2,3),(2,4),(2,5),(3,3)\}$. Let us check this equality for $l=1$ and the rest can be verified similarly.

• •

$y_{0}^{(1)^{\prime}}=\frac{z_{4}^{(2)}z_{4}^{(1)}}{z_{5}^{(2)}z_{5}^{(1)}}=\frac{x_{4}^{(2)}x_{4}^{(1)}}{x_{5}^{(2)}x_{5}^{(1)}}=y_{0}^{(1)}=w_{0}^{(1)}.$

• •

y_{2}^{(1)^{\prime}}=\frac{z_{3}^{(3)}z_{3}^{(2)}z_{3}^{(1)}}{(z_{5}^{(2)})^{2}z_{5}^{(1)}}\big{(}\frac{z_{3}^{(3)}z_{3}^{(2)}}{z_{4}^{(2)}z_{5}^{(2)}}+\frac{z_{4}^{(1)}}{z_{5}^{(2)}}\big{)}^{-1}=\frac{x_{3}^{(3)}x_{3}^{(2)}x_{3}^{(1)}}{(x_{5}^{(2)})^{2}x_{5}^{(1)}}\big{(}\frac{x_{3}^{(3)}x_{3}^{(2)}}{x_{4}^{(2)}x_{5}^{(2)}}+\frac{x_{4}^{(1)}}{x_{5}^{(2)}}\big{)}^{-1}\\ =y_{2}^{(1)}=w_{2}^{(1)}.

• •

y_{3}^{(1)^{\prime}}=\frac{z_{2}^{(2)}z_{2}^{(1)}}{z_{5}^{(2)}}\big{(}\frac{z_{2}^{(2)}z_{3}^{(1)}}{z_{3}^{(3)}}+\frac{z_{3}^{(2)}z_{3}^{(1)}}{z_{5}^{(2)}}+\frac{z_{2}^{(2)}z_{2}^{(1)}}{z_{4}^{(2)}}+\frac{z_{2}^{(2)}z_{4}^{(1)}z_{2}^{(1)}}{z_{3}^{(3)}z_{3}^{(2)}}\big{)}^{-1}\\ =\frac{cx_{2}^{(2)}x_{2}^{(1)}}{x_{5}^{(2)}}\Big{(}\frac{x_{2}^{(2)}x_{3}^{(1)}(cx_{2}^{(2)}x_{2}^{(1)}+x_{3}^{(3)}x_{1}^{(1)})}{x_{3}^{(3)}(x_{2}^{(2)}x_{2}^{(1)}+x_{3}^{(3)}x_{1}^{(1)})}+\frac{x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}}+\frac{cx_{2}^{(2)}x_{2}^{(1)}}{x_{4}^{(2)}}+\frac{cx_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}}\Big{)}^{-1}\\ =y_{3}^{(1)}\cdot\frac{c(y_{3}^{(3)}(y_{3}^{(2)})^{2}y_{3}^{(1)}+y_{2}^{(2)}y_{4}^{(2)}y_{3}^{(2)}y_{3}^{(1)}+y_{2}^{(2)}y_{4}^{(2)}y_{5}^{(1)}y_{2}^{(1)})}{cy_{3}^{(3)}(y_{3}^{(2)})^{2}y_{3}^{(1)}+cy_{2}^{(2)}y_{4}^{(2)}y_{3}^{(2)}y_{3}^{(1)}+y_{2}^{(2)}y_{4}^{(2)}y_{5}^{(1)}y_{2}^{(1)}}=w_{3}^{(1)}.

• •

y_{4}^{(1)^{\prime}}=\big{(}\frac{z_{5}^{(2)}}{z_{3}^{(3)}}+\frac{z_{3}^{(2)}}{z_{2}^{(2)}}+\frac{z_{2}^{(1)}}{z_{1}^{(1)}}\big{)}^{-1}=\big{(}\frac{x_{5}^{(2)}}{x_{3}^{(3)}}+\frac{x_{3}^{(2)}}{x_{2}^{(2)}}+\frac{x_{2}^{(1)}}{x_{1}^{(1)}}\big{)}^{-1}=y_{4}^{(1)}=w_{4}^{(1)}.

• •

$y_{5}^{(1)^{\prime}}=\frac{1}{z_{5}^{(2)}}=\frac{1}{x_{5}^{(2)}}=y_{5}^{(1)}=w_{5}^{(1)}$.

∎

In order to give $\mathcal{V}_{1}$ a $\mathfrak{g}=D_{5}^{(1)}$-geometric crystal structure, we need to define the actions of $e_{0}^{c},\gamma_{0}$, and $\varepsilon_{0}$ on $V_{1}(x)$. We use the $\mathfrak{g}_{1}$-geometric crystal structure on $\mathcal{V}_{2}$ to define the action of $e_{0}^{c}$, $\gamma_{0}$, and $\varepsilon_{0}$ on $V_{1}(x)$ as follows.

[TABLE]

Set $B=B_{x}=\frac{x_{2}^{(2)}x_{3}^{(1)}}{x_{3}^{(3)}}+\frac{x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}},C=C_{x}=\frac{x_{2}^{(2)}x_{2}^{(1)}}{x_{4}^{(2)}}+\frac{x_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}}$ and $A=A_{x}=B_{x}+C_{x}$. The following is one of the main results in this paper.

Theorem 2.4**.**

The algebraic variety $\mathcal{V}=\mathcal{V}_{1}=\{V_{1}(x),e_{k}^{c},\gamma_{k},\varepsilon_{k}\mid k\in I\}$ is a positive geometric crystal for the affine Lie algebra $\mathfrak{g}=D_{5}^{(1)}$ with the $e_{0}^{c}$, $\gamma_{0}$, and $\varepsilon_{0}$ actions on $V_{1}(x)$ given by:

[TABLE]

[TABLE]

Proof.

Since $\mathcal{V}=\mathcal{V}_{1}$ is a positive geometric crystal for $\mathfrak{g}_{0}$, to show that it is a positive geometric crystal for $\mathfrak{g}=D_{5}^{(1)}$ by Definition 1.1, it suffices to show that the following relations involving the [math]-action hold:

1. (1)

$\gamma_{0}(e_{k}^{c}(V_{1}(x)))=c^{a_{k0}}\gamma_{0}(V_{1}(x))$ for all $k\in I$ 2. (2)

$\gamma_{k}(e_{0}^{c}(V_{1}(x)))=c^{a_{0k}}\gamma_{k}(V_{1}(x))$ for all $k\in I$ 3. (3)

$\varepsilon_{0}(e_{0}^{c}(V_{1}(x)))=c^{-1}\varepsilon_{0}(V_{1}(x))$ 4. (4)

$e_{0}^{c_{1}}e_{k}^{c_{2}}=e_{k}^{c_{2}}e_{0}^{c_{1}}$ for all $k\in\{1,3,4,5\}$ 5. (5)

$e_{0}^{c_{1}}e_{2}^{c_{1}c_{2}}e_{0}^{c_{2}}=e_{2}^{c_{2}}e_{0}^{c_{1}c_{2}}e_{2}^{c_{1}}$

Note that  $x_{2}^{(2)^{\prime}}x_{2}^{(1)^{\prime}}=c^{-2}x_{2}^{(2)}x_{2}^{(1)},\ x_{3}^{(3)^{\prime}}x_{3}^{(2)^{\prime}}x_{3}^{(1)^{\prime}}=c^{-2}x_{3}^{(3)}x_{3}^{(2)}x_{3}^{(1)},\ x_{4}^{(2)^{\prime}}x_{4}^{(1)^{\prime}}=c^{-1}x_{4}^{(2)}x_{4}^{(1)},\ x_{5}^{(2)^{\prime}}x_{5}^{(1)^{\prime}}=c^{-1}x_{5}^{(2)}x_{5}^{(1)}$. Relations (1) and (2) follows easily from the defined actions. For example,

[TABLE]

Now we consider relation (3). We have

[TABLE]

since $c(x_{5}^{(1)}+A)B+(cx_{5}^{(1)}+A)C=cx_{5}^{(1)}A+cAB+AC=A(c(x_{5}^{(1)}+B)+C)$. Next we consider relation (4). Since $\mathcal{V}_{2}$ is a $\mathfrak{g}_{1}$-geometric crystal, we have $\bar{e_{5}}^{c_{1}}\bar{e_{2}}^{c_{2}}=\bar{e_{2}}^{c_{2}}\bar{e_{5}}^{c_{1}}$. Then by Proposition 2.3, we have

[TABLE]

Next we show relation (4) for $k=1$. It can be shown for $k=4,5$ similarly. We need to show that

[TABLE]

Set $e_{1}^{c_{2}}(V_{1}(x))=V_{1}(z)$, $e_{0}^{c_{1}}(V_{1}(z))=V_{1}(z^{\prime})$, $e_{0}^{c_{1}}(V_{1}(x))=V_{1}(x^{\prime})$ and $e_{1}^{c_{2}}(V_{1}(x^{\prime}))=V_{1}(u)$. We have to show that

[TABLE]

for $(l,m)\in\{(1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,3)\}$. Observe that $B_{z}=\frac{z_{2}^{(2)}z_{3}^{(1)}}{z_{3}^{(3)}}+\frac{z_{3}^{(2)}z_{3}^{(1)}}{z_{5}^{(2)}}=B_{x}=B$, $C_{z}=\frac{z_{2}^{(2)}z_{2}^{(1)}}{z_{4}^{(2)}}+\frac{z_{2}^{(2)}z_{4}^{(1)}z_{2}^{(1)}}{z_{3}^{(3)}z_{3}^{(2)}}=C_{x}=C$. Hence $A_{z}=A$. Now we have the following.

• •

$z_{1}^{(1)^{\prime}}=\frac{z_{1}^{(1)}}{c_{1}}=\frac{c_{2}x_{1}^{(1)}}{c_{1}}=c_{2}x_{1}^{(1)^{\prime}}=u_{1}^{(1)}$.

• •

$z_{2}^{(1)^{\prime}}=\frac{z_{2}^{(1)}}{c_{1}}=\frac{x_{2}^{(1)}}{c_{1}}=x_{2}^{(1)^{\prime}}=u_{2}^{(1)}$.

• •

$z_{3}^{(1)^{\prime}}=z_{3}^{(1)}\cdot\frac{z_{5}^{(1)}+A_{z}}{c_{1}(z_{5}^{(1)}+B_{z})+C_{z}}=x_{3}^{(1)}\cdot\frac{x_{5}^{(1)}+A}{c_{1}(x_{5}^{(1)}+B)+C}=x_{3}^{(1)^{\prime}}=u_{3}^{(1)}$.

• •

$z_{4}^{(1)^{\prime}}=z_{4}^{(1)}\cdot\frac{z_{5}^{(1)}+A_{z}}{c_{1}(z_{5}^{(1)}+B_{z}+\frac{z_{2}^{(2)}z_{4}^{(1)}z_{2}^{(1)}}{z_{3}^{(3)}z_{3}^{(2)}})+\frac{z_{2}^{(2)}z_{2}^{(1)}}{z_{4}^{(2)}}}=x_{4}^{(1)}\cdot\frac{x_{5}^{(1)}+A}{c_{1}(x_{5}^{(1)}+B+\frac{x_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}})+\frac{x_{2}^{(2)}x_{2}^{(1)}}{x_{4}^{(2)}}}\\ =x_{4}^{(1)^{\prime}}=u_{4}^{(1)}$.

• •

$z_{5}^{(1)^{\prime}}=z_{5}^{(1)}\cdot\frac{z_{5}^{(1)}+A_{z}}{c_{1}z_{5}^{(1)}+A_{z}}=x_{5}^{(1)}\cdot\frac{x_{5}^{(1)}+A}{c_{1}x_{5}^{(1)}+A}=x_{5}^{(1)^{\prime}}=u_{5}^{(1)}$.

• •

$z_{2}^{(2)^{\prime}}=\frac{z_{2}^{(2)}}{c_{1}}=\frac{x_{2}^{(2)}}{c_{1}}=x_{2}^{(2)^{\prime}}=u_{2}^{(2)}$.

• •

$z_{3}^{(2)^{\prime}}=z_{3}^{(2)}\cdot\frac{c_{1}(z_{5}^{(1)}+B_{z})+C_{z}}{c_{1}(c_{1}(z_{5}^{(1)}+\frac{z_{3}^{(2)}z_{3}^{(1)}}{z_{5}^{(2)}})+\frac{z_{2}^{(2)}z_{3}^{(1)}}{z_{3}^{(3)}}+C_{z})}=x_{3}^{(2)}\cdot\frac{c_{1}(x_{5}^{(1)}+B)+C}{c_{1}(c_{1}(x_{5}^{(1)}+\frac{x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}})+\frac{x_{2}^{(2)}x_{3}^{(1)}}{x_{3}^{(3)}}+C)}\\ =x_{3}^{(2)^{\prime}}=u_{3}^{(2)}$.

• •

$z_{4}^{(2)^{\prime}}=z_{4}^{(2)}\cdot\frac{c_{1}(z_{5}^{(1)}+B_{z}+\frac{z_{2}^{(2)}z_{4}^{(1)}z_{2}^{(1)}}{z_{3}^{(3)}z_{3}^{(2)}})+\frac{z_{2}^{(2)}z_{2}^{(1)}}{z_{4}^{(2)}}}{c_{1}(z_{5}^{(1)}+A_{z})}=x_{4}^{(2)}\cdot\frac{c_{1}(x_{5}^{(1)}+B+\frac{x_{2}^{(2)}x_{4}^{(1)}x_{2}^{(1)}}{x_{3}^{(3)}x_{3}^{(2)}})+\frac{x_{2}^{(2)}x_{2}^{(1)}}{x_{4}^{(2)}}}{c_{1}(x_{5}^{(1)}+A)}\\ =x_{4}^{(2)^{\prime}}=u_{4}^{(2)}$.

• •

$z_{5}^{(2)^{\prime}}=z_{5}^{(2)}\cdot\frac{c_{1}z_{5}^{(1)}+A_{z}}{c_{1}(z_{5}^{(1)}+A_{z})}=x_{5}^{(2)}\cdot\frac{c_{1}x_{5}^{(1)}+A}{c_{1}(x_{5}^{(1)}+A)}=x_{5}^{(2)^{\prime}}=u_{5}^{(2)}$.

• •

$z_{3}^{(3)^{\prime}}=z_{3}^{(3)}\cdot\frac{c_{1}(z_{5}^{(1)}+\frac{z_{3}^{(2)}z_{3}^{(1)}}{z_{5}^{(2)}})+\frac{z_{2}^{(2)}z_{3}^{(1)}}{z_{3}^{(3)}}+C_{z}}{c_{1}(z_{5}^{(1)}+A_{z})}=x_{3}^{(3)}\cdot\frac{c_{1}(x_{5}^{(1)}+\frac{x_{3}^{(2)}x_{3}^{(1)}}{x_{5}^{(2)}})+\frac{x_{2}^{(2)}x_{3}^{(1)}}{x_{3}^{(3)}}+C}{c_{1}(x_{5}^{(1)}+A)}\\ =x_{3}^{(3)^{\prime}}=u_{3}^{(3)}$.

Finally to show relation (5) we observe that $\bar{e_{5}}^{c_{1}}\bar{e_{3}}^{c_{1}c_{2}}\bar{e_{5}}^{c_{2}}=\bar{e_{3}}^{c_{2}}\bar{e_{5}}^{c_{1}c_{2}}\bar{e_{3}}^{c_{1}}$ since $\mathcal{V}_{2}$ is a $\mathfrak{g}_{1}$-geometric crystal. Hence by Proposition 2.3, we have

[TABLE]

which completes the proof. ∎

3. Perfect Crystals of type $D_{5}^{(1)}$

For a positive integer $l$, we consider the sets $B^{5,l}$ and $B^{5,\infty}$ as follows.

[TABLE]

For $\mathcal{B}=B^{5,l}\,\text{or}\,B^{5,\infty}$ we define the maps $\tilde{e}_{k},\tilde{f}_{k}:\mathcal{B}\longrightarrow\mathcal{B}\cup\{0\}$, $\varepsilon_{k},\varphi_{k}:\mathcal{B}\longrightarrow\mathbb{Z}$, $0\leq k\leq 5$ and $\text{wt}:\mathcal{B}\longrightarrow P_{cl}$, as follows. First we define conditions $(E_{j}),\ 1\leq j\leq 5$:

[TABLE]

where $(x)_{+}=\text{max}(x,0)$. Then we define conditions $(F_{j})\ (1\leq j\leq 5)$ by replacing $>$ (resp. $\geq$) with $\geq$ (resp. $>$) in $(E_{j})$. Let $b=(b_{ij})\in\mathcal{B}$. Then for $\tilde{e_{k}}(b)=(b^{\prime}_{ij})$ where

[TABLE]

and $b^{\prime}_{ij}=b_{ij}$ otherwise.

Also $\tilde{f_{k}}(b)=(b^{\prime}_{ij})$ where

[TABLE]

and $b^{\prime}_{ij}=b_{ij}$ otherwise.

For $b\in B^{5,l}$ if $\tilde{e}_{k}(b)$ or $\tilde{f}_{k}(b)$ does not belong to $B^{5,l}$, then we assume it to be [math]. The maps $\varepsilon_{k}(b),\ \varphi_{k}(b)$ and $\text{wt}_{k}(b)$ for $k=0,1,2,3,4,5$ are given as follows. We observe that $\text{wt}_{k}(b)=\varphi_{k}(b)-\varepsilon_{k}(b)$, $\varphi(b)=\sum_{k=0}^{5}\varphi_{k}(b)\Lambda_{k}$, $\varepsilon(b)=\sum_{k=0}^{5}\varepsilon_{k}(b)\Lambda_{k}$ and $\text{wt}(b)=\varphi(b)-\varepsilon(b)$.

[TABLE]

Choose elements $b^{0}_{\bf{0}},b^{0}_{\bf{1}},b^{0}_{\bf{2}},b^{0}_{\bf{3}},b^{0}_{\bf{4}},b^{0}_{\bf{5}}$ where

[TABLE]

and $(b^{0}_{\bf{k}})_{ij}=0$  otherwise, for $0\leq k\leq 5$.

As shown in [4], the crystal $B^{5,l}$ is a perfect crystal with the set of minimal elements:

[TABLE]

For $\lambda\in P_{cl}$, consider the crystal $T_{\lambda}=\{t_{\lambda}\}$ with

[TABLE]

$k=0,1,2,3,4,5$. Then for $\lambda,\mu\in P_{cl}$, $T_{\lambda}\otimes B^{5,l}\otimes T_{\mu}$ is a crystal with the structure given by

[TABLE]

where $t_{\lambda}\otimes b\otimes t_{\mu}\in T_{\lambda}\otimes B^{5,l}\otimes T_{\mu}$.

The notion of a coherent family of perfect crystals and its limit is defined in [5]. In the following theorem we prove that the family of $D_{5}^{(1)}$ crystals $\{B^{5,l}\}_{l\geq 1}$ form a coherent family with limit $B^{5,\infty}$ containing the special vector $b^{\infty}={\bf{0}}$ (i.e. $(b^{\infty})_{ij}=0$ for $i\leq j\leq i+4,\ 1\leq i\leq 5$).

Theorem 3.1**.**

*The family of the perfect crystal $\{B^{5,l}\}_{l\geq 1}$ forms a coherent family and the crystal $B^{5,\infty}$ is its limit with the vector $b^{\infty}$. *

Proof.

Set $J=\{(l,b)|l\in\mathbb{Z}_{>0},b\in(B^{5,l})_{\text{min}}\}$. By ([5], Definition 4.1), we need to show that

1. (1)

wt$(b^{\infty})=0,\varepsilon(b^{\infty})=\varphi(b^{\infty})=0,$ 2. (2)

for any $(l,b)\in J$, there exists an embedding of crystals

[TABLE]

where $f_{(l,b)}(t_{\varepsilon(b)}\otimes b\otimes t_{-\varphi(b)})=b^{\infty}$, and 3. (3)

$B^{5,\infty}=\cup_{(l,b)\in J}$ Im $f_{(l,b)}$.

Since $\varepsilon_{k}(b^{\infty})=0,\varphi_{k}(b^{\infty})=0,0\leq k\leq 5$, we have $\varepsilon(b^{\infty})=0,\varphi(b^{\infty})=0$ and hence wt$(b^{\infty})=0$ which proves $(1)$.

Let $l\in\mathbb{Z}_{>0}$ and $b^{0}=(b^{0}_{ij})$ be an element of $(B^{5,l})_{\text{min}}$. Then there exist $a_{k}\in\mathbb{Z}_{\geq 0},0\leq k\leq 5$ such that $a_{0}+a_{1}+2a_{2}+2a_{3}+a_{4}+a_{5}=l$ and

[TABLE]

For any $b=(b_{ij})\in B^{5,l}$ we define a map

[TABLE]

by $f_{(l,b^{0})}(t_{\varepsilon(b^{0})}\otimes b\otimes t_{-\varphi(b^{0})})=b^{\prime}=(b^{\prime}_{ij})$ where $b^{\prime}_{ij}=b_{ij}-b^{0}_{ij}$ for all $i\leq j\leq i+4,\ 1\leq i\leq 5$. Then it is easy to see that

[TABLE]

Hence we have

[TABLE]

For $0\leq k\leq 5,b\in B^{5,l}$, it can be checked easily that the conditions for the action of $\tilde{e}_{k}$ on $b^{\prime}=b-b^{0}$ hold if and only if the conditions for the action of $\tilde{e}_{k}$ on $b$ hold. Hence from the defined action of $\tilde{e}_{k}$, we see that $\tilde{e}_{k}(b^{\prime})=\tilde{e}_{k}(b)-b^{0},0\leq k\leq 5$. This implies that

[TABLE]

[TABLE]

Similarly, we have $f_{(l,b^{0})}(\tilde{f_{k}}(t_{\varepsilon(b^{0})}\otimes b\otimes t_{-\varphi(b^{0})}))=\tilde{f}_{k}(f_{(l,b^{0})}(t_{\varepsilon(b^{0})}\otimes b\otimes t_{-\varphi(b^{0})}))$. Clearly the map $f_{(l,b^{0})}$ is injective with $f_{(l,b^{0})}(t_{\varepsilon(b^{0})}\otimes b^{0}\otimes t_{-\varphi(b^{0})})=b^{\infty}$. This proves (2).

We observe that $\sum_{j=i}^{i+4}b^{\prime}_{ij}=\sum_{j=i}^{i+4}b_{ij}-\sum_{j=i}^{i+4}b^{0}_{ij}=l-l=0$ for all $1\leq i\leq 5$. Also,

[TABLE]

Similarly, we can show that $\sum_{j=i}^{5-t}b^{\prime}_{ij}=\sum_{j=i+t}^{4+t}b^{\prime}_{i+t,j},$ for $2\leq i\leq 4,\ 1\leq t\leq 4$. Hence we have $B^{5,\infty}\supseteq\cup_{(l,b)\in J}$ Im $f_{(l,b)}$. To prove (3) we also need to show that $B^{5,\infty}\subseteq\cup_{(l,b)\in J}$ Im $f_{(l,b)}$. Let $b^{\prime}=(b^{\prime}_{ij})\in B^{5,\infty}$. By (2), we can assume that $b^{\prime}\neq b^{\infty}$. Set

[TABLE]

Let $l=a_{0}+a_{1}+2a_{2}+2a_{3}+a_{4}+a_{5}$. Let $b^{0}=(b^{0}_{ij})$ where

[TABLE]

Then $\varphi(b^{0})=a_{0}\Lambda_{0}+a_{1}\Lambda_{1}+a_{2}\Lambda_{2}+a_{3}\Lambda_{3}+a_{4}\Lambda_{4}+a_{5}\Lambda_{5}$ and $\varepsilon(b^{0})=a_{5}\Lambda_{0}+a_{4}\Lambda_{1}+a_{3}\Lambda_{2}+a_{2}\Lambda_{3}+a_{0}\Lambda_{4}+a_{1}\Lambda_{5}$. It is easy to see that $b^{0}\in(B^{5,l})_{\text{min}}$.

Set $b=(b_{ij})$ where $b_{ij}=b^{\prime}_{ij}+b^{0}_{ij}$. Then $\sum_{j=i}^{i+4}b_{ij}=\sum_{j=i}^{i+4}b^{\prime}_{ij}+\sum_{j=i}^{i+4}b^{0}_{ij}=0+l=l,\ 1\leq i\leq 5$ and we observe that

[TABLE]

Similarly, we can show that $b_{ij}\in\mathbb{Z}_{\geq 0}$ for $i\leq j\leq i+4,\ 2\leq i\leq 5$. We also have

[TABLE]

Similarly, we see that $\sum_{j=i}^{5-t}b_{ij}=\sum_{j=i+t}^{4+t}b_{i+t,j},\ 1\leq i\leq 2,1\leq t\leq 4$.  Also,

[TABLE]

Similarly, $\sum_{j=i}^{t}b_{ij}\geq\sum_{j=i+1}^{t+1}b_{i+1,j},1\leq i<t\leq 4$. Hence $b\in B^{5,l}$.

Then $f_{(l,b^{0})}(t_{\varepsilon(b^{0})}\otimes b\otimes t_{-\varphi(b^{0})})=b^{\prime},$ and $b^{\prime}\in\cup_{(l,b)\in J}$ Im $f_{(l,b)}$ which proves (3). ∎

4. Ultra-discretization of $\mathcal{V}(D_{5}^{(1)})$

It is known that the ultra-discretization of a positive geometric crystal is a Kashiwara crystal [1, 16]. In this section we apply the ultra-discretization functor $\mathcal{UD}$ to the positive geometric crystal $\mathcal{V}=\mathcal{V}(D_{5}^{(1)})$ constructed in Section 2. Then we show that as crystal it is isomorphic to the crystal $B^{5,\infty}$ given in the last section which proves the conjecture in [10] for this case. As a set $\mathcal{X}=\mathcal{UD}(\mathcal{V})=\mathbb{Z}^{10}$. We denote the variables $x_{m}^{(l)}$ in $\mathcal{V}$ by the same notation $x_{m}^{(l)}$ in $\mathcal{UD}(\mathcal{V})=\mathcal{X}$.

Let $x=(x_{4}^{(2)},x_{3}^{(3)},x_{2}^{(2)},x_{5}^{(2)},x_{3}^{(2)},x_{4}^{(1)},x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},x_{5}^{(1)})\in\mathcal{X}$. By applying the ultra-discretization functor $\mathcal{UD}$ to the positive geometric crystal $\mathcal{V}$ in Section 2, we have for $0\leq k\leq 5$:

[TABLE]

We define

[TABLE]

Then we have

[TABLE]

As shown in [1, 16], $\mathcal{X}$ with maps $\tilde{e}_{k},\tilde{f}_{k}:\mathcal{X}\longrightarrow\mathcal{X}\cup\{0\},\;\varepsilon_{k},\varphi_{k}:\mathcal{X}\longrightarrow\mathbb{Z},\;0\leq k\leq 5$ and $\text{wt}:\mathcal{X}\longrightarrow P_{cl}$ is a Kashiwara crystal where for $x\in\mathcal{X}$

[TABLE]

In particular, the explicit actions of $\tilde{f}_{k},1\leq k\leq 5$ on $\mathcal{X}$ is given as follows.

[TABLE]

To determine the explicit action of $\tilde{f_{0}}(x)$ we define conditions $(\breve{F1})-(\breve{F5})$ as follows.

[TABLE]

Then for $x\in\mathcal{X}$ we have $\tilde{f}_{0}(x)=\mathcal{UD}(e_{0}^{c})(x)\arrowvert_{c=-1}$ given by

[TABLE]

Theorem 4.1**.**

The map

[TABLE]

defined by

[TABLE]

is an isomorphism of crystals.

Proof.

First we observe that the map $\Omega^{-1}:\mathcal{X}\rightarrow B^{5,\infty}$ is given by $\Omega^{-1}(x)=b$ where

[TABLE]

Hence the map $\Omega$ is bijective. To prove that $\Omega$ is an isomorphism of crystals we need to show that for $b\in B^{5,\infty}$ and $0\leq k\leq 5$ we have:

[TABLE]

Hence $\varphi_{k}(\Omega(b))=\text{wt}_{k}(\Omega(b))+\varepsilon_{k}(\Omega(b))=\text{wt}_{k}(b)+\varepsilon_{k}(b)=\varphi_{k}(b)$. We observe that the conditions for the action of $\tilde{f}_{k}$ on $\Omega(b)$ in $\mathcal{X}$ hold if and only if the corresponding conditions for the action of $\tilde{f}_{k}$ on $b$ in $B^{5,\infty}$ hold for all $0\leq k\leq 5$. Suppose $\Omega(b)=x$ and $x_{2}^{(2)}+x_{2}^{(1)}>x_{1}^{(1)}+x_{3}^{(2)}$, then $b_{11}+b_{12}+b_{22}>b_{11}+b_{22}+b_{23}$ and $\tilde{f}_{2}(x)=(x_{4}^{(2)},x_{3}^{(3)},x_{2}^{(2)}-1,x_{5}^{(2)},x_{3}^{(2)},x_{4}^{(1)},x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},x_{5}^{(1)})=\Omega(\tilde{f}_{2}(b))$. Similarly, we can show $\tilde{f_{k}}(\Omega(b))=\tilde{f_{k}}(\Omega(b))$ and $\tilde{e_{k}}(\Omega(b))=\tilde{e_{k}}(\Omega(b))$ for $k=0,1,3,4,5$. We also have $\text{wt}_{0}(\Omega(b))=\text{wt}_{0}(x)=-x_{2}^{(2)}-x_{2}^{(1)}=-b_{11}-b_{12}-b_{22}=-b_{11}-b_{12}+b_{23}+b_{24}+b_{25}+b_{26}=\text{wt}_{0}(b)$ for all $b\in B^{5,\infty}$. Similarly, $\text{wt}_{k}(\Omega(b))=\text{wt}_{k}(b)$ for $1\leq k\leq 5$. Also, $\varepsilon_{5}(\Omega(b))=\varepsilon_{5}(x)=\text{max}\{x_{3}^{(3)}-x_{5}^{(2)},x_{3}^{(3)}+x_{3}^{(2)}-2x_{5}^{(2)}+x_{3}^{(1)}-x_{5}^{(1)}\}=\text{max}\{b_{11}+b_{12}+b_{13}-b_{22}-b_{23}-b_{24},b_{11}+b_{12}+b_{13}-b_{22}-b_{23}-2b_{24}+b_{33}-b_{44}\}=\varepsilon_{5}(b)$. Similarly, $\varepsilon_{k}(\Omega(b))=\varepsilon_{k}(b)$ for $0\leq k\leq 4$ which completes the proof. ∎