Norm of $(0,2)$ tensor

Question

If we have a $(0,2)$ tensor $A$ (in my situation, it is actually second fundamental form), I am confused by this notation $|A|$.If we think it as Frobenius norm, write $A$ in local coordinate,i.e., $A=A_{ij}dx^i\otimes dx^j$ where $A_{ij}=A(\partial_i,\partial_j)$, then $$|A|=(\sum A^2_{ij})^{1/2}.$$ But if we think it is a norm induced by metric, then $$|A|=<A,A>^{1/2}=<A_{ij}dx^i\otimes dx^j,A_{mn}dx^m\otimes dx^n>^{1/2}=(g^{im}g^{jn}A_{ij}A_{mn})^{1/2}.$$ It seems that if I choose local orthonormal frame, they are same since $g^{im}=\delta^i_m$. Or the truth is they are same all the time? Any explanation will be helpful.

Answer 1

As mentioned by Andreas the expression

$$|A|=\left(\sum_{i,j=1}^n A_{ij}^2\right)^{1/2}.$$

does not make sense, in general.

Now for a Riemannian manifold $(M,g)$, the Frobenious norm $|A|$ of a $(0,2)$-tensor $A$ is a function of $M$ to $[0,\infty)$, which is given in a coordinate system $p=(x_1,\ldots,x_n)$ by:

$$|A|(p)=\sum_{i_1,i_2,j_1,j_2=1}^n g^{i_1j_1}(p)g^{i_2j_2}(p)A_{i_1j_1}(p)A_{i_2j_2}(p).$$

Given $p\in M$ fixed, you can find a coordinate system $(x_1,\ldots,x_n)$ around $p$ such that $g^{ij}(p)=\delta_{ij}$, namely any normal coordinates centered at $p$. Observe that for $q\neq p$ in the domain of the normal coordinates, we have $g^{ij}(q)\neq \delta_{ij}$ in general. Nonetheless, in these normal coordinates we obtain the following expression of the Frobenious norm of $A$ at $p$:

$$|A|(p) = \left(\sum_{i,j=1}^n A_{ij}^2(p)\right)^{1/2}.$$

With this observation, the first expression makes sense pointwise, i.e. you can assume that $|A|=(\sum A^2_{ij})^{1/2}$ as functions over $M$.

Also observe that given an $g(p)$-orthonormal basis $\{e_1,\ldots,e_n\}$ of $T_p M$ there are cannonical normal coordinates $(x_1,\ldots,x_n)$ centered at $p$, such that $\partial_i(p)=e_i$ (see Chapter 2, Gray, A., "Tubes", second edition, Birkhäuser Verlag, Basel (2004) for references about normal coordinates).

Answer 2

If you are just talking about a smooth manifold, a $(0,2)$-tensor field does not have a well-defined norm. The Frobenius norm of a local coordinate expression depends on the choice of coordinates and thus has no geometric meaning.

In the presence of a metric (which probably is the case you are thinking about) the second definition is independent of the choice of coordinates and thus is a well-defined norm of a $(0,2)$-tensor field. This does coincide with the Frobenius norm of the components in a local orthonormal frame, but these are not local coordinate expressions, since orthonormal coordinates only exist if the metric is flat. Moreover, the transformation between different local orthonormal frames (smooth functions to the orthogonal group) is completely different from the transformation between the componente in coordinates (deriviatives of chart-changes).