# The proof of a Riemannian metric as a $(0,2)$ tensor

I was told that the Riemannian metric is a $(0,2)$ tensor. I have trouble understand this.

I know very little geometry, I learnt that the Weingarten map of a hypersurface is a linear map from $T_x\Sigma$ to $T_x$ for each $x$ and so define a $(1,1)$-tensor field on $\Sigma$. The idea is the following: for each $u\in T_x\Sigma$ and $\omega\in T_x^*\Sigma$, apply first $\mathcal{W}$ on $u$ (i.e.$\mathcal{W}$ acting on $u$) to give a new tangent vector, then apply the element of the dual space $T_x^*\Sigma$: $\omega$ on $\mathcal{W}$ to produce a real number. Essentially, $\omega$ is a continuous linear functional acting on $T_x\Sigma$.

However, I have no idea how to apply the idea to show a Riemannian metric is a $(0,2)$ tensor.

Could anyone help for a detail answer in extending what I understood (written above)? Thanks.

• "I have trouble understanding this". One way to understand what a (n,m) tensor is through the way it transforms under a change of coordinates. Ex.: a (1,0) (also 1-cotra-variant) tensor is a usual vector, a (0,1) (also 1-covariant) tensor is a form (vector of the dual space), a (1,1) is a normal "matrix", to fix ideas. Is that something you knew?
– MASL
Commented Sep 11, 2015 at 0:22
• Notice then that the metrics acts on vectors, i.e., a metric $G$ acts on two vectors $u, v$ and gives a number $G(u,v)\in \mathbb{R}$.
– MASL
Commented Sep 11, 2015 at 0:24
• @MASL Looking at your first comment and trying to explain what I wrote in another way...on my notes it says "1 form is a $(1,0)$ tensor-field" while a smooth vector filed is a $(0,1)$ tensor-field. I doubt this is obviously false? and back to this case, first, the $(0,1)$ tensor gives a vector field, then apply the 1-form, or the convector or $(1,0)$ tensor to get a new tangent vector... Commented Sep 11, 2015 at 2:12
• @MASL Looking at your 2nd comments, if, intuitively,it is like...if our $\omega$ acts one vector we get the 1-form, if $\omega$ act on two vectors we get 2-form, or the $(0,2)$ tensor..So the $G$ plays the role of $\omega$, right ? Commented Sep 11, 2015 at 2:19
• Have a look at the wikipedia page on tensors, in particular en.wikipedia.org/wiki/Tensor#Examples I just want to point you to the "simple" view in terms of components: The Riemannian metric $g_{\mu\nu}$, a $2-$covariant tensor, acts on two vectors (1-contravariant tensors) $u^\mu\,v^\nu$ giving a number $g_{\mu\nu}u^\mu v^\nu$ (a 0 order tensor).
– MASL
Commented Sep 11, 2015 at 2:33

The Shape Operator $S: T_pM \rightarrow T_pM$ this is true. In particular, this is calculated in terms of dot-products of the coordinate velocities and the normal vector field to the surface; $S(\alpha') \cdot \alpha' = \alpha'' \cdot U$ where $U$ is the normal vector field to $M$. Some usual notation: (ala O'neill's Elementary Differential Geometry) $$S(X_u) \cdot X_u = L, \ \ S(X_u) \cdot X_v = M, \ \ S(X_v) \cdot X_v = N$$ where $(u,v) \mapsto X(u,v)$ is the patch on $M$. Or, $$S(X_u) \cdot X_u = U\cdot X_{uu} = L, \\ S(X_u) \cdot X_v = U\cdot X_{uv} = M, \\ S(X_v) \cdot X_v = U\cdot X_{vv} = N.$$ Anyway, clearly the shape operator is very much tied to the particulars of the normal vector field. Hence the name "shape operator". We learn that the eigenvalues of the shape operator are the principal curvatures. Moreover, the Gaussian curvature is $K = \text{det}(S)$ and the mean curvature $H = \frac{1}{2}\text{trace}(S)$.
In contrast, the first fundamental form $I = Edu \otimes du+Fdu \otimes dv +Gdv \otimes dv$ (aka metric) has coefficients which are simply based on: $$E = X_u \cdot X_u, \ \ \ F = X_u \cdot X_v, \ \ \ G = X_v \cdot X_v$$ It turns out we can also calculate $K$ from $E,F,G$, see page 5 of my notes based on O'neill's Section 6.6. In other words, while $K$ appears to involve the shape of the surface, it is in fact an illusion of the initial set-up of the Weingarten map. The intrinsic geometry of the surface is not based on the shape operator, but, rather those geometric concepts which are based on the distances and angles as derived from the metric alone. There are surfaces which share the same intrinsic geometry yet have radically different shapes. See this animation of the helicoid morphing to catenoid. You see, the helicoid and catenoid share the same warping functions $E,F,G$ hence the same intrinsic geometry (but obviously not the same normal vector field and hence "shape")
So, the shape operator and metric are related, but, the relation is not as direct as you might expect. From the viewpoint of a given surface with a given patch, both are built from expressions derived from coordinate velocities (the metric) and accelerations (the Shape operator). It is clear from the formulas I give that the tensor type of the metric is not the same as that of the shape operator. The fact that the metric is a $(0,2)$ tensor is manifest from the formula: $$I = E du \otimes du +Fdu \otimes dv+ G dv \otimes dv.$$ For example $du \otimes dv(X,Y) = du(X)dv(Y)$ which is linear in both $X$ and $Y$ and the output is a scalar. $I$ is built from a sum of these fundamental type $(0,2)$ tensors and is hence a $(0,2)$ tensor. Sorry if I haven't quite answered your question, but, I believe this to be your true question.