# Hamilton's Proof of the Tensor Maximum Principle

My questions come from Richard Hamilton's Three-Manifolds with Positive Ricci Curvature paper. I'm trying to work through parts of the paper so I can better understand the Ricci Flow for my research. I'm working in section 9 on Preserving Positive Ricci Curvature, the proof of Theorem 9.1.

I'm really having a hard time writing down the precise details of the bound $|\tilde{N}_{ij}-N_{ij}|\le C|\tilde{M}_{ij}-M_{ij}|$ where $C$ depends only on $\max(|\tilde{M}_{ij}|+|M_{ij}|)$. It somehow seems "obvious" from the fact that $N$ is a polynomial, which is the way Hamilton defines it, until you get around to writing down what $N$ looks like in terms of contractions of products of $M$, which is a horrific mess.

I think the best effort I have is choosing normal coordinates at a point such that the metric is $\delta_{ij}$ and $M_{ij}$ is diagonalized, as it's symmetric. Then the norm on the left starts to look like $|Q(\tilde{\lambda_i})-Q(\lambda_i)|$ where $Q$ is a polynomial of $n$ variables, by $Q(\lambda_i)$ I mean $Q(\lambda_1,...,\lambda_n)$, and $\tilde{\lambda_i}=\lambda_i+\epsilon(\delta+t)$ are the perturbed eigenvalues. Then using the mean-value inequality for multivariable functions, for some $0<c<1$ this is $$\le |\nabla Q(\lambda_i+c\epsilon(\delta+t))|(\sqrt{n}\epsilon(\delta+t))=|\nabla Q(\lambda_i+c\epsilon(\delta+t))||\tilde{M}_{ij}-M_{ij}|$$ taking into account the coordinates in which we are working. I am not sure how this $|\nabla Q|$ would depend on $(|\tilde{M}_{ij}|+|M_{ij}|)$ at this point, if it does at all, or if Hamilton had something different in mind.

I also got stuck again immediately after that, where he concludes, apparently from the estimate above, that $N_{ij}v^iv^j\ge-C\epsilon\delta$ with $C$ depending only on $\max|M_{ij}|$. I guess it's not clear to me that an estimate on the norm of a tensor has anything to do with it's action on vectors, given that this action is defined regardless of the presence of a metric.

Any constructive thoughts will be much appreciated.

Since I don't have enough reputation to comment, I just wanna say something here. Notice that the null vector has unit length. So $|(\tilde{N_{ij}}-N_{ij})v^iv^j|\le |(\tilde{N_{ij}}-N_{ij})||v|^2=|(\tilde{N_{ij}}-N_{ij})|\le C\epsilon\delta$ and C depends only on $Max |M_{ij}|$ for enough small $\epsilon$ and $\delta$. So, $N_{ij}v^iv^j-\tilde{N_{ij}}v^iv^j\ge -C\epsilon\delta$, Null vector condition implies $\tilde{N_{ij}}v^iv^j\ge 0$, then the argument follows.