# What is the modulus of a tensor on a Riemannian 3-manifold?

Let $v^i$ be a vector on a Riemannian 3-manifold with metric $g_{ij}$ embedded inside a 3+1 space-time such that for some constant $N_M$ it satisfies the inequality $g_{ij}v^iv^j \leq N_M ^2$. Let $K$ be a symmetric rank-2 tensor on the 3-manifold. Then apparently the following holds:

$$\vert K_{ij} v^i v^j \vert \leq \vert K \vert _g N_M ^2.$$

This looks like some sort of a Cauchy-Schwarz inequality but given that $K$ is a tensor as described, I don't understand what the notation on the RHS means. For a rank-2 symmetric tensor $K$ what does $\vert K \vert _g$ mean?

If one knows that for some function $N$, $g_{ij}v^iv^j \leq N^2,$ where the function $N$ is itself bounded between constants

$$N_m \leq N \leq N_M,$$

then using inequalities like the above one can apparently show the following bound:

$$\int _{t_1} ^t \frac{1}{N} \Big(-v^i \partial _i N - \frac{dN}{dt} + K_{ij}v^iv^j\Big) dt \leq -2\log N_m + \frac{1}{N_m} \int _{t_1}^t (\vert \nabla N \vert _g N_M + \vert K \vert _g N_M ^2 )dt,$$

for some fixed $t_1$ and $t$.

I can't understand that first "log" term in the above.

Also once the above bound is shown does it follow that the integral can be unbounded above or below depending solely on the property of the function $N$? If yes then what would be needed of $N$ to make the integral unbounded above or below?

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I believe $|K|_g$ is the induced norm. –  KennyTM Sep 24 '10 at 17:51
@Kenny could you elaborate a little more? Like if you could add in a few lines of the derivation. –  Anirbit Sep 24 '10 at 18:05
It's just a speculation because I remembered a similar thing in real analysis. I don't have a derivation. Or probably $|K|_g$ is just defined as $\max_{v\ne 0}\frac{|K_{ij}v^iv^j|}{g_{ij}v^iv^j}$. –  KennyTM Sep 24 '10 at 18:10
I think the $|K|_g$ here comes from the following fact: An inner product on V induces an inner product on all tensors. In fact an inner product on $V$ gives a natural isomorphism between $V$ and $V^*$ by $v\rightarrow \langle v,\cdot\rangle$ (I'm assuming $V$ is finite dimensional here). Then, by declaring this isomorphism to be an isometry, we get an inner product on $V*$. If $e_i$ is an o.n. basis of $V$, then $\langle e_i, \cdot \rangle$ is a basis of $V*$. Now by declaring things of the form $e_i\otimes e_j^*$, etc to be orthonormal, one gets an inner product on all tensors. –  Jason DeVito Sep 24 '10 at 20:58
I don't know why I never answered the second part of your question. The $\log$ term comes from using the fundamental theorem of calculus, and the other terms come from using Cauchy-Schwartz with the induced norm I mentioned in my answer and what is known in geometric analysis as Kato's inequality, $|\nabla |T|| \le |\nabla T|$. –  Glen Wheeler May 2 '11 at 8:52

I am just going to answer the first question. For a two-tensor we have

$$|T|^2 = \left< T,T \right> = g^{ik}g^{jl}T_{ij}T_{kl},$$

where $g$ is the metric, and the summation convention is understood. Note that this is the standard inner product structure induced by $g$, as mentioned by Jason DeVito. You might like to try and prove that this really does define a norm in the Riemannian case.

More generally, for a $(k,l)$ ($k$ times contravariant, $l$ times covariant) tensor field $T$ (on the manifold) we have

$$|T|^2 = \left< T,T \right> = g^{j_1q_1}g^{j_2q_2}\cdots g^{j_lq_l}g_{i_1p_1}g_{i_2p_2}\cdots g_{i_kp_k}T^{i_1i_2\cdots i_k}_{j_1j_2\cdots j_l}T^{p_1p_2\cdots p_k}_{q_1q_2\cdots q_l}.$$

For your other questions, you could do worse than look in 'Einstein Manifolds' by Besse.

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