# Definitions for Norm of a Tensor

I am reading a book on differential geometry and at one point the following statement is made:

$$|\text{Ric} + \text{Hess}(f)|^2 \geq \frac{1}{n}|R+\Delta f|^2.$$

The justification is that 'we have estimated the norm of the symmetric tensor $$\text{Ric} + \text{Hess}(f)$$ in terms of its trace'. I know a bit about matrix norms and norms of bounded linear operators, but I am curious what exactly is the norm of an arbitrary tensor.

Is it just the Frobenius or $$l^2$$ norm for an arbitrary tensor? So if it is rank 4, you take the tensor and contract with itself over the 4 indices and take the square root of the result, but then where is the $$1/n$$ coming from? I know $$n$$ would be the dimension of the space, but why do we need to divide by this and why is the Frobenius norm always bigger than or equal to the trace (the sum of the diagonal elements in matrix terms).

Let us consider norms on function spaces before discussing tensors on manifolds. For a subset $$U\subset\mathbb{R}^n$$, recall that there are many different normed spaces, all of whose elements are functions on $$U$$. For example, we have the spaces $$L^p(U)$$ for $$1\leq p\leq\infty$$ and the spaces $$C^k(U)$$ for $$k\in\mathbb{N}$$. Note that every $$L^p$$ norm takes into account only the pointwise values of a function, while the $$C^k$$ norms (as well as various Sobolev norms) are defined in means of a function and its derivatives (up to some order).

Analogously, there are many different norms on tensor spaces. Even though you did not specify in your post which norm the author uses (maybe the author did not specify that either), I get the impression this is a norm which takes into account only the pointwise values of tensors and ignores their derivatives. This is because the inequality in question actually holds pointwise, and to see this, we just need some basic linear algebra.

So let $$(V,g)$$ be a (finite dimensional) real inner product space, and let $$B$$ be a symmetric bilinear form on $$V$$. A simple (even if not the most canonical) way to define the (Hilbert-Schmidt) norm of $$B$$ is as follows. Let $$e_1,\ldots,e_n\in V$$ be an orthonormal basis, and let $$f^1,\ldots,f^n\in V^*$$ denote the dual basis. Express $$B$$ in means of this basis, that is, write $$B=B_{ij}f^i\otimes f^j.$$ Then the norm of $$B$$ is given by $$\|B\|^2=\sum_{i,j}|B_{ij}|^2.$$ Now, the inner product identifies bilinear forms with endomorphisms, and so, $$B$$ can be thought of as a self-adjoint endomorphism with eigenvalues $$\lambda_1,\ldots,\lambda_n$$. The norm of $$B$$, as defined above, then satisfies $$\|B\|^2=\sum|\lambda_i^2|,$$ while the trace of $$B$$ is just the sum of the $$\lambda$$'s. Finally, the desired inequality now reads $$\sum\lambda_i^2\ge\frac {1}{n}\left(\sum\lambda_i\right)^2,$$which follows directly from the Cauchy-Schwarz inequality.

Edit: By the Cauchy-Schwarz inequality, \begin{align}\left(\sum\lambda_i\right)^2&=\langle(\lambda_1,\ldots,\lambda_n),(1,\ldots,1)\rangle^2\\&\le\|(\lambda_1,\ldots,\lambda_n)\|_2^2\cdot\|(1,\ldots,1)\|_2^2\\&=n\sum\lambda_i^2\;.\end{align}

• Apologies, why does the $1/n$ factor follow from Cauchy-Schwarz?
– Tom
Commented Feb 28, 2019 at 19:51
• @Tom I edited my answer, please take a look. Commented Feb 28, 2019 at 21:30
• Do you mind if I ask another question, so if we define $B$ in the way that you define it, why does it follow that $\parallel B \parallel ^2 = \Sigma |\lambda_i^2|$? Is that a general result for self-adjoint operators/maps?
– Tom
Commented Feb 28, 2019 at 22:48
• @Tom Yeah, it is a general result which basically follows from the fact that every symmetric matrix is orthogonally diagonalizable (and the fact that the norm is well-defined). Commented Feb 28, 2019 at 23:18