The law of sine of triangle ($\triangle ABC$) in the Euclidean plane is well known as follows

$\begin{eqnarray} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{abc}{2S}, \end{eqnarray}$

(1.1)

where $S=\sqrt{p(p-a)(p-b)(p-c)}$, $p=\frac{1}{2}(a+b+c)$.

Let $a,b,c$ be the edge-lengths of a triangle $ABC$ in the hyperbolic space with curvature $-1$. Then we have the law of sine of hyperbolic triangle $ABC$ as follows (see [1])

$\begin{eqnarray} \frac{\sinh a}{\sin A}=\frac{\sinh b}{\sin B}=\frac{\sinh c}{\sin C}=\frac{\sinh a\sinh b\sinh c}{2\Delta}, \end{eqnarray}$

(1.2)

where $\Delta=\sqrt{\sinh p(\sinh p-\sinh a)(\sinh p-\sinh b)(\sinh p-\sinh c)}$, $p=\frac{1}{2}(a+b+c)$.

We denote by $a, b, c$ the edge-lengths of a triangle $ABC$ in the spherical space with curvature $1$. Then we have the law of sine of spherical triangle $ABC$ as follows (see [2])

$\begin{eqnarray} \frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}=\frac{\sin a\sin b\sin c}{2\Delta}, \end{eqnarray}$

(1.3)

where $\Delta=\sqrt{\sin p(\sin p-\sin a)(\sin p-\sin b)(\sin p-\sin c)}$, $p=\frac{1}{2}(a+b+c)\in(0,\pi)$.

The law of sine of triangle in Euclidean plane were generalized to the $n$-dimensional simplex in $n$-dimensional Euclidean space $E^n$. Let $\{A_{0},A_{1},\cdots, A_{n}\}$ be the vertex sets of $n$-dimensional simplex $\Omega_{n}(E)$ in the $n$-dimensional Euclidean $E^n$. Denote by $V$ the volume of the simplex $\Omega_{n}(E)$, and $F_{i}$ $(i=0,1,\cdots,n)$ the areas of $i$-th face $f_{i}=\{A_{0},$ $A_{1},$ $\cdots,$ $ A_{i-1},$ $A_{i+1},$ $ \cdots, $ $A_{n}\}$ ($(n-1)$-dimensional simplex) of the simplex $\Omega_{n}(E)$. In 1968, Bators defined the $n$-dimensional sines of the $n$-dimensional vertex angles $\alpha_{i}$ $(i=0,1,\cdots,n)$ for the $n$-dimensional simplex $\Omega_{n}(E)$, and established the law of sines for $\Omega_{n}(E)$ as follows (see [3])

$\begin{eqnarray} \frac{F_{0}}{\sin\alpha_{0}}=\frac{F_{1}}{\sin\alpha_{1}}=\cdots=\frac{(n-1)!\prod_{j=0}^{n}F_{j}}{(n V)^{n-1}}. \end{eqnarray}$

(1.4)

Obviously, formula $(1.4)$ is generalization of formula $(1.1)$ in $n$-dimensional Euclidean space $E^n$. Then, some different forms of generalization about formula $(1.1)$ was given in [4, 5, 6].

From 1970s, many geometry researchers were attempted to generalize formulas $(1.2)$ and $(1.3)$ to an $n$-dimensional hyperbolic simplex (spherical simplex), to establish the law of sines in $n$-dimensional hyperbolic space $H^n$ and in $n$-dimensional spherical space $S^n$. In 1978, Erikson defined the $n$-dimensional polar sine of $i$-th face $f_{i}(P_{i}\notin f_{i})$ of $\Omega_{n}(S)$ in $S_{n,1}$ (see [7]) as follows

$\begin{eqnarray} ^{n}{\rm Pol}\sin F_{i}=|[\nu_{0},\nu_{1},\cdots,\nu_{i-1},\nu_{i+1},\cdots,\nu_{n}]|\quad (i=0,1,\cdots,n). \end{eqnarray}$

(1.5)

Let $^{n}\sin P_{i}$ be the $n$-dimensional sine of the $i$-th angle of $\Omega_{n}(S)$ (see [7]). The law of sines in the $n$-dimensional spherical space $S_{n,1}$ was obtained in [7] as follows

$\begin{eqnarray} \frac{^{n}{\rm Pol}\sin F_{0}}{^{n}\sin P_{0}}=\frac{^{n}{\rm Pol}\sin F_{1}}{^{n}\sin P_{1}}=\cdots=\frac{^{n}{\rm Pol}\sin F_{n}}{^{n}\sin P_{n}}. \end{eqnarray}$

(1.6)

In 1980s, Yang and Zhang (see [8, 9, 10]) made a large number of basic works of geometric inequality in $n$-dimensional hyperbolic space $H^n$ and in $n$-dimensional spherical space $S^n$, and established the law of cosines in $n$-dimensional hyperbolic space $H^n$ and $n$-dimensional spherical space $S^n$. But they did not establish the law of sines in $n$-dimensional hyperbolic space $H^n$ and $n$-dimensional spherical space $S^n$. In addition, some new geometric inequality about "metric addition" involving volume and $n$-dimensional space angle of simplex in $H^{n}(K)$ and $S^{n}(K)$ is established.

In this paper, we give generalizations of $(1.2)$ and $(1.3)$ in $n$-dimensional hyperbolic space $H^n$ and in $n$-dimensional spherical space $S^n$, and establish the law of sines $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^n$ and $n$-dimensional spherical space $S^n$. As their applications, we obtain Veljan-Korchmaros type inequalities and Hadamard type inequalities in $n$-dimensional hyperbolic space $H^{n}$ and $n$-dimensional spherical space $S^{n}$.

We consider the model of the hyperbolic space in Euclidean space (see [9]).

Let $B$ be a set whose elements $x(x_{1},x_{2},\cdots,x_{n})$ are in an $n$-dimensional vector space and meet the following condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}<1$. Given a distance between $x$ and $y$ in the set $B$, denote by $xy$ satisfying

$\begin{eqnarray*}\cosh\sqrt{-K}xy=\frac{1-x_{1}y_{1}-x_{2}y_{2}-\cdots -x_{n}y_{n}}{\sqrt{1-x_{1}^{2}-x_{2}^{2}-\cdots -x_{n}^{2}}\sqrt{1-y_{1}^{2}- y_{2}^{2}-\cdots -y_{n}^{2}}}.\end{eqnarray*}$

Then the metric space with this distance in $R^{n+1}$ is called $n$-dimensional hyperbolic space with curvature $K (<0)$, denote by $H^{n}(K)$.

Let $\Sigma_{n}(H)$ be an $n$-dimensional simplex in the $n$-dimensional hyperbolic space $H^{n}(K)$, and $\{P_{0},P_{1},\cdots,P_{n}\}$ be its vertexes, $a_{ij}$ $(0\leqslant i,j\leqslant n)$ be its edge-length, $V$ be its volume, respectively.

To give the law of sines in $n$-dimensional hyperbolic space $H^{n}(K)$, we give the following definition.

Definition 2.1 Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$. $n$ edges $P_{0}P_{i}$ $(i=1,2,\cdots,n)$ with initial point $P_{0}$ form an $n$-dimensional space angle $P_{0}$ of the simplex $\Sigma_{n}(H)$. Let $\widehat{i,j}$ be the angle formed by two edges $P_{0}P_{i}$ and $P_{0}P_{j}$. The sine of the $n$-dimensional space angle $P_{0}$ of the simplex $\Sigma_{n}(H)$ is defined as follows

$\begin{eqnarray} \sin P_{0}=(\det Q_{0})^{\frac{1}{2}}, \end{eqnarray}$

(2.1)

where

$\begin{equation*} Q_{0}=\left[ \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right]\quad (i,j=1,2,\cdots,n). \end{equation*}$

Similarly, we can define the sine of the $n$-dimensional space angle $P_{i}$ $(i=1,2,\cdots,n)$ of the simplex $\Sigma_{n}(H)$.

At first, we prove that this definition is sensible.

Actually, for $n$-ray $P_{0}P_{i}$ $(i=1,2,\cdots,n)$ of the simplex $\Sigma_{n}(H)$ in $n$-dimensional hyperbolic space $H^{n}(K)$, and denote by $\widehat{i,j}$ $(i,j=1,2,\cdots,n)$ the included angle between two rays $P_{0}P_{i}$ and $P_{0}P_{j}$. According to [12], we know that there exist $n$-ray $P'_{0}P'_{i}$ $(i=1,2,\cdots,n)$ which are independence in $n$-dimensional Euclidean space $E^{n}$, such that the included angle between two rays $P'_{0}P'_{i}$ and $P'_{0}P'_{j}$ is also $\widehat{i,j}$ $(i,j=1,2,\cdots,n)$. Assume that $\overrightarrow{\alpha_{i}}$ $(i=1,2,\cdots,n)$ denote the unit vector of the vector $\overrightarrow{P'_{0}P'_{i}}$ $(i,j=1,2,\cdots,n)$, then the unit vectors $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$ are also independence. So the Gram matrix $Q_{0}$ of the unit vectors $\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$ is positive and it is easy to know that $0<\det Q_{0}\leqslant 1$. Therefore, this definition is sensible.

Remark Especially two-dimensional space angle of two-dimensional hyperbolic simplex (that is hyperbolic triangle) is just interior angle of hyperbolic triangle. So the $n$-dimensional space angle of $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$ is extension of interior angle of hyperbolic triangle.

Definition 2.2 (see [12]) Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$, its volume $V$ is the real number satisfying

$\begin{eqnarray} \sinh^{2}\sqrt{-K}V=\frac{(-1)^{n}}{2^{n}\cdot(n!)^{2}}\det(\Lambda_{n}(H)), \end{eqnarray}$

(2.2)

where $\Lambda_{n}(H)=(\cosh\sqrt{-K}a_{ij})_{i,j=0}^{n}$.

Theorem 2.1 Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in the $n$-dimensional hyperbolic space $H^{n}(K)$, we have

$\begin{eqnarray} &&\frac{\prod\limits_{1\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}}{\sin P_{0}}=\frac{\prod\limits_{0\leqslant i<j\leqslant n,i,j\neq 1}\sinh\sqrt{-K}a_{ij}}{\sin P_{1}}=\cdots \nonumber\\ &=&\frac{\prod\limits_{0\leqslant i<j\leqslant n-1}\sinh\sqrt{-K}a_{ij}}{\sin P_{n}}=\frac{\prod\limits_{0\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}}{2^{\frac{n}{2}}\cdot n!\cdot \sinh\sqrt{-K}V}, \end{eqnarray}$

(2.3)

where definitions of $\sin P_{i}(i=0,1,\cdots)$ are the same as Definition $2.1$.

Remark When $n=2$ in Theorem $2.1$, it is the law of sine of a hyperbolic triangle.

Lemma 2.2 (see [1]) (the law of cosine of a hyperbolic triangle) For hyperbolic triangle $ABC$ in $H^{2}(-1)$, then

$\begin{eqnarray} \cosh a=\cosh b\cdot\cosh c-\sinh b\cdot\sinh c\cdot \cos A, \end{eqnarray}$

(2.4)

where $a,b,c$ be edge-lengths of hyperbolic triangle $ABC$ and $A$ be the interior angle.

Lemma 2.3 Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$, we have

$\begin{eqnarray} \sinh^{2}\sqrt{-K}V=\frac{1}{2^{n}\cdot (n!)^{2}}\Big(\prod_{i=1}^{n}\sinh^{2}\sqrt{-K}a_{0i}\Big)\cdot \left| \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right|, \end{eqnarray}$

(2.5)

where $\widehat{i,j}$ $(i,j=1,2,\cdots, n)$ be the included angle between the edges $P_{0}P_{i}$ and $P_{0}P_{j}$.

Proof Assume that the row or the column number of determinant $\det(\Lambda_{n}(H))$ begins from $0$. Now, we transform the determinant $\det(\Lambda_{n}(H))$ as follows:

(1) for $i=1,2,\cdots,n$, plus $(-\cosh\sqrt{-K}a_{0i})$ times the $0$-th row to the $i$-th row;

(2) expanding the determinant at the $0$-th column;

(3) for $i=1,2,\cdots,n$, $\frac{1}{(\sinh\sqrt{-K}a_{0i})}$ times the $i$-th row and $\frac{1}{(\sinh\sqrt{-K}a_{0i})}$ times the $i$-th column.

$\begin{align*} &\det(\Lambda_{n}(H))= \left| \begin{array}{cccc} 1 & \cosh\sqrt{-K}a_{01} &\cdots & \cosh\sqrt{-K}a_{0n}\\ \cosh\sqrt{-K}a_{10} & 1 &\cdots & \cosh\sqrt{-K}a_{1n}\\ \cdots&\cdots & \cdots &\cdots \\ \cosh\sqrt{-K}a_{n0}&\cosh\sqrt{-K}a_{n1} &\cdots &1 \end{array} \right|\\ =&\small\left| \begin{array}{cccc} 1 & \cosh\sqrt{-K}a_{01} &\cdots & \cosh\sqrt{-K}a_{0n}\\ 0 & 1-\cosh^{2}\sqrt{-K}a_{01} &\cdots & \cosh\sqrt{-K}a_{1n}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01}\\ \cdots &\cdots & \cdots &\cdots \\ 0&\cosh\sqrt{-K}a_{n1}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01} &\cdots &1-\cosh^{2}\sqrt{-K}a_{0n} \end{array} \right| \end{align*}$

$\begin{align*} =&\small\left| \begin{array}{ccc} 1-\cosh^{2}\sqrt{-K}a_{01} & \cdots & \cosh\sqrt{-K}a_{1n}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01}\\ \cdots &\cdots & \cdots\\ \cosh\sqrt{-K}a_{n1}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01} &\cdots &1-\cosh^{2}\sqrt{-K}a_{0n} \end{array} \right| \end{align*}$

$\begin{align*} =&\Big(\prod_{i=1}^{n}\sinh^{2}\sqrt{-K}a_{0i}\Big)\small\left| \begin{array}{ccc} \frac{1-\cosh^{2}\sqrt{-K}a_{01}}{\sinh^{2}\sqrt{-K}a_{01}} & \cdots & \frac{\cosh\sqrt{-K}a_{1n}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01}} {\sinh\sqrt{-K}a_{01}\sinh\sqrt{-K}a_{0n}}\\ \cdots &\cdots &\cdots \\ \frac{\cosh\sqrt{-K}a_{n1}-\cosh\sqrt{-K}a_{0n}\cosh\sqrt{-K}a_{01}}{\sinh\sqrt{-K}a_{01}\sinh\sqrt{-K}a_{0n}} &\cdots &\frac{1-\cosh^{2}\sqrt{-K}a_{0n}}{\sinh^{2}\sqrt{-K}a_{0n}} \end{array} \right|\\ =&(-1)^{n}\Big(\prod_{i=1}^{n}\sinh^{2}\sqrt{-K}a_{0i}\Big)\left| \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right|. \end{align*}$

By $(2.3)$, we have

$\begin{eqnarray} \det(\Lambda_{n}(H))=\frac{2^{n}\cdot(n!)^{2}}{(-1)^{n}}\sinh^{2}\sqrt{-K}V. \end{eqnarray}$

(2.6)

Substituting $(2.6)$ into above equality, we get $(2.5)$.

Proof of Theorem 2.1 According to Definition $2.1$, equality $(2.5)$ may be written as

$\begin{eqnarray} \sinh\sqrt{-K}V=\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\Big(\prod_{i=1}^{n}\sinh\sqrt{-K}a_{0i}\Big)\cdot\sin P_{0}. \end{eqnarray}$

(2.7)

Now we only prove that

$\begin{eqnarray} \frac{\prod\limits_{1\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}}{\sin P_{0}}=\frac{\prod\limits_{0\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}}{2^{\frac{n}{2}}\cdot n!\cdot \sinh\sqrt{-K}V}. \end{eqnarray}$

(2.8)

Applying $(2.7)$, we get

$\begin{align*} \Big(\prod\limits_{1\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}\Big)\cdot \sinh\sqrt{-K}V =&\Big(\prod\limits_{1\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{ij}\Big)\cdot\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\prod_{i=1}^{n} \sinh\sqrt{-K}a_{0i}\cdot\sin P_{0}\\ =&\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\prod_{0\leqslant i<j\leqslant n}\sinh\sqrt{-K}a_{0i}\cdot\sin P_{0}. \end{align*}$

From above equality, we obtain $(2.8)$.

Similarly, we can prove that other equalities in $(2.3)$ also hold. The proof of Theorem $2.1$ is completed.

We consider the model of a spherical space in the Euclidean space (see [6]): the distance $xy$ between two points $x$ and $y$ in the points set $S=\{x(x_{1},x_{2},\cdots,x_{n+1}):x^{2}_{1}+x^{2}_{2}+\cdots+x^{2}_{n+1}=\frac{1}{K},K>0 \text{is constant number}\}$ in the $n+1$-dimensional Euclidean space $E^{n+1}$ is the minimal non-negative real number satisfying

$\begin{eqnarray} \cos\sqrt{K}xy=\frac{x_1y_1+x_2y_2+\cdots+x_{n+1}y_{n+1}}{\sqrt{x_1^2+x_2^2+\cdots+ x_{n+1}^2}\sqrt{y_1^2+y_2^2+\cdots+y_{n+1}^2}}. \end{eqnarray}$

(3.1)

The metric space with this distance in the point set $S$ is called $n$-dimensional spherical space with curvature $K>0$, and denote by $S_{n}(K)$.

Let $\Omega_{n}(S)$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $S^{n}(K)$, and $\{A_{0},A_{1},\cdots,A_{n}\}$ be its vertexes, $a_{ij}$ $(0\leqslant i,j\leqslant n)$ be its edge-length, $V$ be its volume.

To give the law of sines in $n$-dimensional spherical space $S^{n}(K)$, we give the following definition.

Definition 3.1 Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, $n$ edges $A_{0}A_{i}$ $(i=1,2,\cdots,n)$ with initial point $A_{0}$ form an $n$-dimensional space angle $A_{0}$ of the simplex $\Omega_{n}(A)$. Denote by $\widehat{i,j}$ be the included angle between two edges $A_{0}A_{i}$ and $A_{0}A_{j}$. The sine of the $n$-dimensional space angle $A_{0}$ of the simplex $\Omega_{n}(S)$ is defined as follows

$\begin{eqnarray} \sin A_{0}=(\det B_{0})^{\frac{1}{2}}, \end{eqnarray}$

(3.2)

where

$\begin{equation*} B_{0}=\left[ \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right]\quad (i,j=1,2,\cdots,n). \end{equation*}$

Similarly, we can define the sine of the $n$-dimensional space angle $A_{i}$ $(i=1,2,\cdots,n)$ of the simplex $\Omega_{n}(S)$.

At first, we prove that this definition is sensible.

Actually, $\widehat{i,j}$ be the included angle between unit tangent vector $\vec{e_{i}}$ and $\vec{e_{j}}$ at point $A_{0}$ of two arcs $\overset\frown{A_{0}A_{i}}$ and $\overset\frown{A_{0}A_{j}}$. Because the unit tangent vectors $\vec{e_{1}},\vec{e_{2}},\cdots,\vec{e_{n}}$ are independence, the Gram matrix $B_{0}$ of the unit vectors $\vec{e_{1}},\vec{e_{2}},\cdots,\vec{e_{n}}$ is positive, and it is easy to know that $0<\det B_{0}\leqslant 1$. Therefore, this definition is sensible.

Remark Especially two-dimensional space angle of two-dimensional spherical simplex (that is spherical triangle) is just interior angle of spherical triangle. So the $n$-dimensional space angle of $n$-dimensional simplex in $n$-dimensional spherical space $H^{n}(K)$ is extension of interior angle of spherical triangle.

Definition 3.2 (see [11]) Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, its volume $V$ is the minimal non-negative real number satisfying

$\begin{eqnarray} \sin^{2}\sqrt{K}V=\frac{1}{2^{n}\cdot n!^{2}}({\rm det}\Lambda_n), \end{eqnarray}$

(3.3)

where $\Lambda_{n}(S)=(\cos\sqrt{K}a_{ij})_{i,j=0}^{n}$.

Theorem 3.1 Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, we have

$\begin{eqnarray} \cfrac{\prod\limits_{1\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}}{\sin A_{0}}=\cfrac{\prod\limits_{\substack{0\leqslant i<j\leqslant n\\i,j\neq 1}}\sin\sqrt{K}a_{ij}}{\sin A_{1}}=\cdots=\cfrac{\prod\limits_{0\leqslant i<j\leqslant n-1}\sin\sqrt{K}a_{ij}}{\sin A_{n}}=\cfrac{\prod\limits_{0\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}}{2^{\frac{n}{2}}\cdot n!\cdot \sin\sqrt{K}V}, \end{eqnarray}$

(3.4)

where definitions of $\sin A_{i}(i=0,1,\cdots)$ are the same as Definition $3.1$.

Remark When $n=2$ in Theorem $3.1$, it is the law of sine of a spherical triangle.

Lemma 3.2 (see [2]) (the law of cosine of a spherical triangle) For spherical triangle $ABC$ in $S^{2}(1)$, then

$\begin{eqnarray} \cos a=\cos b\cdot\cosh c+\sin b\cdot\sin c\cdot \cos A, \end{eqnarray}$

(3.5)

where $a,b,c$ be edge-lengths of spherical triangle $ABC$ and $A$ be interior angle.

Lemma 3.3 Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, we have

$\begin{eqnarray} \sin^{2}\sqrt{K}V=\frac{1}{2^{n}\cdot (n!)^{2}}\Big(\prod_{i=1}^{n}\sin^{2}\sqrt{K}a_{0i}\Big)\cdot \left| \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right|, \end{eqnarray}$

(3.6)

where $\widehat{i,j}$ $(i,j=1,2,\cdots, n)$ be the included angle between the edges $P_{0}P_{i}$ and $P_{0}P_{j}$.

Proof Assume that the row or the column number of determinant $\det(\Lambda_{n}(S))$ begins from $0$. Now, we transform the determinant $\det(\Lambda_{n}(S))$ as follows:

(1) for $i=1,2,\cdots,n$, plus $(-\cos\sqrt{K}a_{0i})$ times the $0$-th row to the $i$-th row;

(2) expanding the determinant at the $0$-th column;

(3) for $i=1,2,\cdots,n$, $\frac{1}{(\sin\sqrt{K}a_{0i})}$ times the $i$-th row and $\frac{1}{(\sin\sqrt{K}a_{0i})}$ times the $i$-th column.

$\begin{align*} &\det(\Lambda_{n}(S))= \left| \begin{array}{cccc} 1 & \cos\sqrt{K}a_{01} &\cdots & \cos\sqrt{K}a_{0n}\\ \cos\sqrt{K}a_{10} & 1 &\cdots & \cos\sqrt{K}a_{1n}\\ &\cdots & \cdots & \\ \cos\sqrt{K}a_{n0}&\cos\sqrt{K}a_{n1} &\cdots &1 \end{array} \right|\\ =&\left| \begin{array}{cccc} 1 & \cos\sqrt{K}a_{01} &\cdots & \cos\sqrt{K}a_{0n}\\ 0 & 1-\cos^{2}\sqrt{K}a_{01} &\cdots & \cos\sqrt{K}a_{1n}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01}\\ &\cdots & \cdots &\cdots \\ 0&\cos\sqrt{K}a_{n1}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01} &\cdots &1-\cos^{2}\sqrt{K}a_{0n} \end{array} \right| \end{align*}$

$\begin{align*} =&\left| \begin{array}{ccc} 1-\cos^{2}\sqrt{K}a_{01} & \cdots & \cos\sqrt{K}a_{1n}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01}\\ \cdots&\cdots &\cdots \\ \cos\sqrt{K}a_{n1}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01} &\cdots &1-\cos^{2}\sqrt{K}a_{0n} \end{array} \right|\end{align*}$

$\begin{align*} =&\Big(\prod_{i=1}^{n}\sin^{2}\sqrt{K}a_{0i}\Big)\left| \begin{array}{ccc} \frac{1-\cos^{2}\sqrt{K}a_{01}}{\sin^{2}\sqrt{K}a_{01}} & \cdots & \frac{\cos\sqrt{K}a_{1n}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01}} {\sin\sqrt{K}a_{01}\sin\sqrt{K}a_{0n}}\\ \cdots&\cdots & \cdots \\ \frac{\cos\sqrt{K}a_{n1}-\cos\sqrt{K}a_{0n}\cos\sqrt{K}a_{01}}{\sin\sqrt{K}a_{01}\sin\sqrt{K}a_{0n}} &\cdots &\frac{1-\cos^{2}\sqrt{K}a_{0n}}{\sin^{2}\sqrt{K}a_{0n}} \end{array} \right|\\ =&\Big(\prod_{i=1}^{n}\sin^{2}\sqrt{K}a_{0i}\Big)\left| \begin{array}{cccc} 1 & & & \cos\widehat{i,j}\\ & 1 & & \\ & & \ddots & \\ \cos\widehat{i,j}& & &1 \end{array} \right|. \end{align*}$

By$(3.3)$ we have

$\begin{eqnarray} \det(\Lambda_{n}(S))=2^{n}\cdot(n!)^{2}\sin^{2}\sqrt{K}V. \end{eqnarray}$

(3.7)

Substituting $(3.7)$ into above equality, we get $(3.6)$.

Proof of Theorem 3.1 According to Definition $3.1$, equality $(3.6)$ may be written as

$\begin{eqnarray} \sin\sqrt{K}V=\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\Big(\prod_{i=1}^{n}\sin\sqrt{K}a_{0i}\Big)\cdot\sin A_{0}. \end{eqnarray}$

(3.8)

Now we only prove that

$\begin{eqnarray} \frac{\prod\limits_{1\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}}{\sin A_{0}}=\frac{\prod\limits_{0\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}}{2^{\frac{n}{2}}\cdot n!\cdot \sin\sqrt{K}V}. \end{eqnarray}$

(3.9)

Applying $(3.8)$, we get

$\begin{align*} \big(\prod\limits_{1\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}\big)\cdot \sin\sqrt{K}V=&\big(\prod\limits_{1\leqslant i<j\leqslant n}\sin\sqrt{K}a_{ij}\big)\cdot\frac{1}{2^{\frac{n}{2}}\cdot n!}\prod_{i=1}^{n}\sin\sqrt{K}a_{0i}\cdot\sin A_{0}\\ =&\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\big(\prod_{0\leqslant i<j\leqslant n}\sin\sqrt{K}a_{0i}\big)\cdot\sin A_{0}. \end{align*}$

From above equality, we obtain $(3.9)$.

Similarly, we can prove that other equalities in $(3.4)$ also hold. The proof of Theorem $3.1$ is completed.

On basis of Section $2$ and Section $3$, we are easy to establish Veljan-Korchmaros type inequalities and Hadamard type inequalities in the $n$-dimensional hyperbolic space $H^{n}(K)$ and the $n$-dimensional spherical space $S^{n}(K)$. In addition, some new geometric inequality about "metric addition" [11, 14] involving Volume and $n$-dimensional angle of simplex in $H^{n}(K)$ and $S^{n}(K)$ is established.

Theorem 4.1 Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$, we have

$\begin{eqnarray} \prod_{0\leq i<j\leq n}\sinh\sqrt{-K} a_{ij}\geqslant 2^{\frac{n(n+1)}{2}}(n!)^{n+1}(\sinh\sqrt{-K}V)^{n+1}. \end{eqnarray}$

(4.1)

Proof  Because $\sin P_{i}\leqslant 1$ in $(2.3)$, we have

$\begin{eqnarray} \sinh^{2}\sqrt{-K}V\leqslant\frac{1}{2^{n}\cdot (n!)^{2}}\prod_{j=1}^{n}\sinh\sqrt{-K}a_{ij},\quad j=0,1,\cdots,n. \end{eqnarray}$

(4.2)

Multiplying by above those inequalities for $j=0,1,\cdots,n$, we get $(4.1)$.

Theorem 4.2 Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, we have

$\begin{eqnarray} \prod_{0\leqslant i<j\leqslant n}\sin\sqrt{K} a_{ij}\geqslant 2^{\frac{n(n+1)}{2}}(n!)^{n+1}(\sin\sqrt{K}V)^{n+1}. \end{eqnarray}$

(4.3)

Proof  Because $\sin A_{i}\leqslant 1$ in $(3.3)$, we have

$\begin{eqnarray} \sin^{2}\sqrt{K}V\leqslant\frac{1}{2^{n}\cdot (n!)^{2}}\prod_{j=1}^{n}\sin\sqrt{K}a_{ij},\quad j=0,1,\cdots,n. \end{eqnarray}$

(4.4)

Multiplying by above those inequalities for $j=0,1,\cdots,n$, we get $(4.3)$.

Since $\sin P_{0}\leqslant 1$ in $(2.7)$ and $\sin A_{0}\leqslant 1$ in $(3.7)$, thus we obtain Hadamard type inequalities in $n$-dimensional hyperbolic space $H^{n}(K)$ and $n$-dimensional spherical space $S^{n}(K)$ as follows:

Theorem 4.3 Suppose that $\Sigma_{n}(H)=\{P_{0},P_{1},\cdots,P_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional hyperbolic space $H^{n}(K)$, we have

$\begin{eqnarray} \sinh\sqrt{-K}V\leqslant\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\prod_{i=1}^{n}\sinh\sqrt{-K}a_{0i}. \end{eqnarray}$

(4.5)

Theorem 4.4 Suppose that $\Omega_{n}(S)=\{A_{0},A_{1},\cdots,A_{n}\}$ be an $n$-dimensional simplex in $n$-dimensional spherical space $S^{n}(K)$, we have

$\begin{eqnarray} \sin\sqrt{K}V\leqslant\frac{1}{2^{\frac{n}{2}}\cdot(n!)}\prod_{i=1}^{n}\sin\sqrt{K}a_{0i}. \end{eqnarray}$

(4.6)

Theorem 4.5 Let $\Sigma''_{n}(H)$ be $n$-dimensional metric addition simplex which is formed two $n$-dimensional simplexes $\Sigma_{n}(H)$ and $\Sigma'_{n}(H)$ by "metric addition" operation in $n$-dimensional hyperbolic space $H^{n}(K)$, we have

$\begin{eqnarray} \big(\frac{\sinh\sqrt{-K}V''}{\sin P''_{i}}\big)^{\frac{1}{n}}\geqslant \big(\frac{\sinh\sqrt{-K}V}{\sin P_{i}}\big)^{\frac{1}{n}}+ \big(\frac{\sinh\sqrt{-K}V'}{\sin P'_{i}}\big)^{\frac{1}{n}}\quad(i=0,1,\cdots,n). \end{eqnarray}$

(4.7)

Equality obtain if and only if the simplex $\Sigma_{n}(H)$ and $\Sigma'_{n}(H)$ is regular.

Theorem 4.6  Let $\Omega''_{n}(S)$ be $n$-dimensional metric addition simplex which is formed two $n$-dimensional simplexes $\Omega_{n}(S)$ and $\Omega'_{n}(S)$ by "metric addition" operation in $n$-dimensional spherical space $S^{n}(K)$, we have

$\begin{eqnarray} \big(\frac{\sin\sqrt{K}V''}{\sin A''_{i}}\big)^{\frac{1}{n}}\geqslant \big(\frac{\sin\sqrt{K}V}{\sin A_{i}}\big)^{\frac{1}{n}}+ \big(\frac{\sin\sqrt{K}V'}{\sin A'_{i}}\big)^{\frac{1}{n}}\quad(i=0,1,\cdots,n). \end{eqnarray}$

(4.8)

Equality obtain if and only if the simplex $\Omega_{n}(S)$ and $\Omega'_{n}(S)$ is regular.

Lemma 4.7 (see [15]) Let $a_{k},b_{k}\ge 0$, then

$\begin{eqnarray}\prod_{k=1}^{n}(a_{k}+b_{k})^{\frac{1}{n}}\ge \big(\prod_{k=1}^{n}a_{k}\big)^{\frac{1}{n}}+\big(\prod_{k=1}^{n}b_{k}\big)^{\frac{1}{n}}. \end{eqnarray}$

(4.9)

The Proof of Theorem $4.5$ According to the definition of "metric addition" [11] in $H^{n}(K)$, we have $\sinh\sqrt{-K} a''_{0i}=\sinh\sqrt{-K}a_{0i}+\sinh\sqrt{-K}a_{0i}(\text{for} i=1,2,\cdots,n).$ Thus

$\prod_{i=1}^{n}\sinh\sqrt{-K}a''_{0i}=\prod_{i=1}^{n}(\sinh\sqrt{-K}a_{0i}+\sinh^{2}\sqrt{-K}a'_{0i})$

$\frac{1}{n}$-th power on the both sides, we get

$\begin{eqnarray} \big(\prod_{i=1}^{n}\sinh\sqrt{-K}a''_{0i}\big)^{\frac{1}{n}}=\big[\prod_{i=1}^{n} (\sinh\sqrt{-K}a_{0i}+\sinh\sqrt{-K}a'_{0i})\big]^{\frac{1}{n}}. \end{eqnarray}$

(4.10)

By $(4.9)$ and $(4.10)$, we have

$\begin{align} \nonumber \big(\prod_{i=1}^{n}\sinh\sqrt{-K}a''_{0i}\big)^{\frac{1}{n}}&=\big[\prod_{i=1}^{n}(\sinh\sqrt{-K}a_{0i} +\sinh\sqrt{-K}a'_{0i})\big]^{\frac{1}{n}}\\ &\geqslant\big(\prod_{i=1}^{n}\sinh\sqrt{-K}a_{0i}\big)^{\frac{1}{n}}+\big(\prod_{i=1}^{n} \sinh\sqrt{-K}a'_{0i}\big)^{\frac{1}{n}}. \end{align}$

(4.11)

By $(2.3)$, we obtain

$\big(\frac{\sinh\sqrt{-K}V''}{\sin P''_{0}}\big)^{\frac{2}{n}}\geqslant\big(\frac{\sinh\sqrt{-K}V}{\sin P_{0}}\big)^{\frac{2}{n}} +\big(\frac{\sinh\sqrt{-K}V'}{\sin P'_{0}}\big)^{\frac{2}{n}}.$

Similarly, inequality $(4.7)$ is easy proved for $i=1,2,\cdots,n$.

The proof of Theorem $4.6$ is the same as the proof of Theorem $4.5$.