So I have more questions coming from Hamilton's "Three-Manifolds with Positive Ricci Curvature" paper. 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.
First I'll discuss the bound $|\tilde{N}_{ij}-N_{ij}|\le C|\tilde{M}_{ij}-M_{ij}|$, where $C$ depends only on $max(|\tilde{M}_{ij}|+|M_{ij}|)$.
Fix a point $x\in X$ and coordinates such that at $x$, $g_{ij}=\delta_{ij}$ and $M_{ij}$ is diagonal with eigenvalues $\lambda_1,...,\lambda_n$, which is possible since $M$ is a symmetric $2$-tensor and thus can be orthogonally diagonalized. From the definition of $N$ as a polynomial $p(M,g)$ formed by contracting products of $M$ with $g$, we see that $N$ is diagonal at $x$ and the $i$th diagonal entry is $q_i(\lambda_1,...,\lambda_n)$ where the $q_i$ are polynomials. Let $\tilde{\lambda_i}=\lambda_i+\epsilon(\delta+t)$ be the perturbed eigenvalues. We have
$|\tilde{N}_{ij}-N_{ij}|^2=(q_1(\tilde{\lambda_i})-q_1(\lambda_i))^2+\ldots+(q_n(\tilde{\lambda_i})-q_n(\lambda_i))^2$.
Now
$|q_k(\tilde{\lambda_i})-q_k(\lambda_i)|\le |\nabla q_k((1-c)\lambda_i+c\tilde{\lambda_i})||\tilde{\lambda_i}-\lambda_i|=|\nabla q_k((1-c)\lambda_i+c\tilde{\lambda_i})||\tilde M_{ij}-M_{ij}|$
for some $0\le c\le 1$ by the Mean Value Inequality. The quantity $|\nabla q_k|$ is a polynomial since $q_k$ is a polynomial, so just by applying the triangle inequality and using that $c\le 1$, $|\nabla q_k((1-c)\lambda_i+c\tilde{\lambda_i})|$ is bounded by a polynomial in $|\tilde{\lambda_k}|+|\lambda_k|$ where $k$ is the index maximizing $|\tilde{\lambda_k}|+|\lambda_k|$, and $|\tilde{\lambda_k}|+|\lambda_k|$ is straightforwardly bounded by $|\tilde{M}_{ij}|+|M_{ij}|$. Thus
$|\nabla q_k((1-c)\lambda_i+c\tilde{\lambda_i})|\le C_k$, where $C_k$ depends on $max(|\tilde{M}_{ij}|+|M_{ij}|)$.
Then it follows that
$|\tilde{N}_{ij}-N_{ij}|=[(q_1(\tilde{\lambda_i})-q_1(\lambda_i))^2+\ldots+(q_n(\tilde{\lambda_i})-q_n(\lambda_i))^2]^{1/2}\le (C_1+\ldots+C_n)^{1/2}|\tilde M_{ij}-M_{ij}|=C|\tilde M_{ij}-M_{ij}|$.
There's also the claim that $N_{ij}v^iv^j\ge -C\epsilon\delta$ to address. First we need several simple facts which are straightforward to verify.
$\langle T_{i_1\ldots i_a},S_{j_1\ldots j_a}\rangle= T_{i_1\ldots i_a}S^{i_1\ldots i_a}$ where $S_{i_1\ldots i_a}=g_{i_1j_1}\ldots g_{i_aj_a}S^{j_1\ldots j_a}$
$\langle T_{i_1\ldots i_a},T_{j_1\ldots j_a}\rangle=\langle T^{i_1\ldots i_a},T^{j_1\ldots j_a}\rangle$, so in particular $|v|=|v^{\flat}|$
$|T\otimes S|=|T||S|$
From this it then follows that
$|(\tilde{N}_{ij}-N_{ij})v^iv^j|=\langle \tilde{N}-N, v^{\flat}\otimes v^{\flat}\rangle\le |\tilde{N}_{ij}-N_{ij}||v^{\flat}\otimes v^{\flat}|=|\tilde{N}_{ij}-N_{ij}||v^{\flat}|^2=|\tilde{N}_{ij}-N_{ij}||v|^2=|\tilde{N}_{ij}-N_{ij}|\le C|\tilde{M}_{ij}-M_{ij}|=C\epsilon(\delta+t)\sqrt{n}\le 2C\epsilon\delta\sqrt{n}$
since $0\le t\le T$
by using the facts above, the Cauchy-Schwarz inequality, and the bound obtained in the first part.
Then $2C\epsilon\delta\sqrt{n}\ge |(\tilde{N}_{ij}-N_{ij})v^iv^j|\ge (\tilde{N}_{ij}-N_{ij})v^iv^j\ge -N_{ij}v^iv^j$ by the null-eigenvector condition, from which it follows that $N_{ij}v^iv^j\ge -C\epsilon\delta$, where this $C$ is just a constant multiple of our previous $C$ from above.
It is easy to check that $|\tilde{M}|^2=|M|^2+2\epsilon(\delta+t)\langle M,g\rangle +{\epsilon}^2(\delta+t)^2n\le (|M|+2\sqrt{n})^2$ so we can make our constant depend only on $max |M_{ij}|$ by keeping $\epsilon, \delta\le 1$ as Hamilton states.