## Newton tensor

Suppose ${A:V\rightarrow V}$ is a symmetric endomorphism of vector space ${V}$, ${\sigma_k}$ is the ${k-}$th elementary symmetric function of the eigenvalue of ${A}$. Then

$\displaystyle \det(A+tI)=\sum_{k=0}^n \sigma_k t^{n-k}$

One can define the ${k-}$th Newton transformation as the following

$\displaystyle \det(A+tI)(A+tI)^{-1}=\sum_{k=0}^{n-1}T_k(A)t^{n-k-1}$

This means

$\displaystyle \det(A+tI)=\sum_{k=0}^{n-1}T_k(A+tI)t^{n-k-1}$

$\displaystyle =T_0 t^n+\sum_{k=0}^{n-2}(A\cdot T_k(A)+T_{k+1}(A))t^{n-k-1}+T_{n-1}(A)$

By comparing coefficients of ${t}$, we get the relations of ${T_k}$

$\displaystyle T_0=1,\quad A\cdot T_k(A)+T_{k+1}(A)=\sigma_{k+1}I,\, 0\leq k\leq n-2\quad T_{n-1}(A)=\sigma_n$

Induction shows

$\displaystyle T_{k}(A)=\sigma_kI-\sigma_{k-1}A+\cdots+(-1)^kA^k$

For example

$\displaystyle T_1(A)=\sigma_1I-A$

$\displaystyle T_2(A)=\sigma_2-\sigma_1A+A^2$

One of the important property of Newton transformation is that: Suppose ${F(A)=\sigma_k(A)}$, then

$\displaystyle F^{ij}=\frac{\partial F}{\partial A_{ij}}=T_{k-1}^{ij}(A)$

The is because

$\displaystyle \frac{\partial }{\partial A_{ij}}\det(A+tI)=\det(A+tI)((A+tI)^{-1})_{ij}.$

If ${A\in \Gamma_k}$, then ${T_{k-1}(A)}$ is positive definite and therefore ${F}$ is elliptic.

Remark: Hu, Z., Li, H. and Simon, U. . Schouten curvature functions on locally conformally flat Riemannian manifolds. Journal of Geometry, 88(1${-}$2), (2008), 75${-}$100.