## A matrix inequality and its geometric application

In this post I will show a matrix inequality by Huang and Wu, and give one geometric application of it. See their paper for other applications.

 Proposition 1 (Huang-Wu) For an ${n\times n}$ matrix, ${A=(a_{ij})}$, $\displaystyle \sigma_1 \widetilde \sigma_1 = \sigma_2 +\frac{n}{2(n-1)} \widetilde \sigma_1 ^2+\sum_{1\leq i where ${\displaystyle\sigma_1 = \sum_{i=1}^n a_{ii}}$, ${\displaystyle\widetilde \sigma_1 = \sum_ {i=2}^n a_{ii}}$ and ${\displaystyle\sigma_2= \sum_{1\leq i. In particular if ${A}$ is real and ${\displaystyle\sum_{1\leq i, $\displaystyle \sigma_1 \widetilde \sigma_1 \geq \sigma_2 +\frac{n}{2(n-1)} \widetilde \sigma_1 ^2 .$

Proof: Denote ${a_{ii}}$ by ${\lambda_i}$. Consider

$\displaystyle \begin{array}{rcl} \sigma_1\widetilde \sigma_1&=& (\lambda_1+\cdots +\lambda_n)(\lambda_2+\cdots+ \lambda_n)\\&=& \lambda_1(\lambda_2+\cdots +\lambda_n)+\widetilde \sigma_1^2\\ &=& \displaystyle\sigma_2 -\sum_{2\leq i

We want to eliminate ${\displaystyle\sum_{2\leq i in the above. This is done by observing (summation runs from ${2\leq i if not specified)

$\displaystyle \begin{array}{rcl} \sum(\lambda_i-\lambda_j)^2&=&\displaystyle\sum (\lambda _i^2 + \lambda _j^2) -2\sum\lambda_i\lambda_j\\ &=&\displaystyle(n-2)\sum_{j\geq 2}\lambda_i^2-2\sum\lambda_i\lambda_j\\ &=&\displaystyle (n-2)(\sum_{j\geq 2}\lambda_j)^2-2(n-1)\sum \lambda_i\lambda_j. \end{array}$

Substitute this expression of ${\sum \lambda_i\lambda_j}$ into the first equation, we are done. $\Box$

Let ${(M^n,g)}$ be a Riemannian manifold and ${\Sigma}$ be a ${(n-1)}$-dimensional submanifold of ${M}$. Let ${e_1,\cdots, e_n}$ be a local orthnormal frame on ${M}$ such that ${e_2,\cdots ,e_n}$ are tangential to ${\Sigma}$. Denote by ${R}$ and ${Rc=(R_{ij})}$ the scalar curvature and the Ricci curvature of ${M}$ respectively, ${H_\Sigma}$ and ${A_\Sigma}$ to be the mean curvature and the shape operator of ${\Sigma}$ in ${M}$ respectively.

 Proposition 2 With the notations and assumptions above, $\displaystyle \begin{array}{rcl} \displaystyle R (R_\Sigma -H_\Sigma ^2 +|A_\Sigma|^2)&=& \displaystyle\frac{1}{2} (R^2-|Rc|^2 )+ \frac{n}{2(n-1)}(\sum_{i=2}^nR_{ii})^2+ \sum_{1\leq i

Proof: The result follows by applying Proposition 1 to ${(R_{ij})}$ directly. Note that by the Gauss equation ${R_\Sigma= \sum_{i=2}^n R_{ii}+H_\Sigma^2-|A_\Sigma|^2}$,

$\displaystyle \widetilde \sigma_1(R_{ij})= R_\Sigma - H_\Sigma ^2+|A_\Sigma|^2.$

Also,

$\displaystyle \sigma _2(R_{ij})= \sum_{1\leq i

$\Box$

 Corollary 3 With the notations and assumptions above, if ${R\geq |Rc|}$ on ${M}$, then $\displaystyle R_\Sigma \geq H_\Sigma ^2-|A_\Sigma |^2.$ The equality holds if and only if at that point, ${R=|Rc(e_1,e_1)|}$ and ${R_{ij}=0}$ if ${(i,j)\neq (1,1)}$.

Proof: Clearly the inequality holds at any point ${p\in \Sigma}$ where ${R>0}$. Suppose ${R(p)=0}$, then by Proposition 2 we have ${\sum_{i=2}^n R_{ii}=0}$ at ${p}$. But the Gauss equation shows that ${\sum_{i=2}^n R_{ii}= R_\Sigma - H_\Sigma ^2+|A_\Sigma |^2}$. Hence the inequality holds.

The necessary and sufficient condition for the equality can be easily seen from the RHS of the equation in Proposition 2. $\Box$

When ${M^n}$ is a hypersurface in ${\mathbb{R}^{n+1}}$, then the condition in Corollary 3 can be replaced by the extrinsic curvature conditions of ${M}$. Let ${(\lambda_1, \cdots, \lambda_n)}$ be the principal curvatures of ${M}$ in ${\mathbb{R}^{n+1}}$. We define the ${k}$-th mean curvature of ${M}$ to be

$\displaystyle \sigma _k(M)=\sum_{1\leq i_1<\cdots

For even ${k}$, ${\sigma_k(M)}$ is an intrinsic quantity (by Reilly) depending only on the induced metric of ${M}$. For odd ${k}$, ${\sigma_k(M)}$ is well-defined up to a sign (i.e. the choice of the normal ${\nu}$), but ${\sigma_{k-1}(M)\sigma _{k+1}(M)}$ is independent of the choice of ${\nu}$.

 Proposition 4 With the same notations as before, if ${R}$ is nonnegative and $\displaystyle \sigma_2(M)^2+\sigma _1(M)\sigma _3(M)+ 2\sigma _4(M)\geq 0, \ \ \ \ \ (1)$ then ${R\geq |Rc|}$ on ${M}$. In particular, in this case, ${R_\Sigma \geq H_\Sigma ^2-|A_\Sigma |^2}$.

Proof: To simplify notations, we simply denote ${\sigma _k(M)}$ by ${\sigma_k}$. There is an orthonormal ${\{e_i\}_{i=1}^n}$ which diagonalizes the shape operator: ${Ae_i =\lambda _i e_i}$. Then ${Rc}$ is diagonalized by ${e_i}$ (using Gauss equation) with

$\displaystyle R_{ii} = \lambda _i \sum_{j\neq i}\lambda _j = \sigma _1 \lambda _i -\lambda _i^2.$

Thus

$\displaystyle |Rc|^2 = \sum _i (\sigma_1 \lambda _i - \lambda _i ^2)^2= \sigma_1^2 \sum _i \lambda _i^2-2\sigma _1 \sum_i \lambda _i^3 + \sum _i \lambda _i^4. \ \ \ \ \ (2)$

Let ${p_k = \sum _{i=1}^n \lambda_i^k}$. Then by Newton’s identities, we have

$\displaystyle \begin{array}{rcl} \begin{cases} p_1=\sigma _1,\\ p_2=\sigma_1 p_1-2\sigma_2=\sigma _1^2-2\sigma_2,\\ p_3= \sigma _1 p_2 -\sigma_2 p_1 + 3\sigma_3=\sigma_1 ^3 -3\sigma_1\sigma_2 +3\sigma_3,\\ p_4 =\sigma _1 p_3 -\sigma_2 p_2 + \sigma _3 p_1 -4 \sigma _4= \sigma_1^4 - 4 \sigma_1 ^2\sigma _2 +4\sigma_1\sigma_3 +2\sigma_2^2 -4\sigma_4.\end{cases} \end{array}$

Plugging these into (2), and using ${R=2\sigma_2}$, we have

$\displaystyle R^2-|Rc|^2= 2(\sigma_2^2 +\sigma_1\sigma_3 +2 \sigma_4).$

The result follows. $\Box$

 Remark 1 The condition (1) in Proposition 4 is automatically satisfied when ${n=2}$ (${\sigma_k }$ is defined to be zero if ${k>n}$). However the inequality is trivial in this case, as ${\Sigma}$ is a curve, ${R_\Sigma=0}$ and ${|H_\Sigma |=|A_\Sigma|}$ is just the geodesic curvature of ${\Sigma}$. In general, (1) can also be replaced by other stronger condition. For example, by a Maclaurin-type inequality (see Hardy, Littlewood, Polya’s Inequalities p.52 Theorem 53, or here for a slightly stronger statement) $\displaystyle \sigma_{k-1}\sigma_{k+1}\leq \sigma_k ^2,$ so we can replace the condition by ${\sigma_1(M)\sigma _3 (M)+\sigma _4(M)\geq 0}$.