Is a Riemannian metric a $2$-form?

Question

In Lee's Riemannian Manifolds; An introduction to Curvature, he defines a Riemannian metric as an element of $\Gamma(T^2_0M)$, a $(2,0)$-tensor. Is this the same thing as a $2$-form? Is there a difference between being a section of $T^\ast M\times T^\ast M$ and a section of $T^2_0M$? A difference between a $(2,0)$-tensor and $2$-form?

Answer 1

There is a rather significant difference. I'm not sure how comfortable you are with various vector bundle constructions, but both 2-forms and metrics are special elements of $E = T^*M \otimes T^*M$ the second dual tensor power bundle, which is identified with the space of smooth varying bilinear maps (that is to say a section $q$ of the bundle $E$ is a bilinear map on each tangent space such that $q(X,Y)$ is smooth for every vector fields $X,Y$. 2-forms are the space of $q$ such that $q(X,Y) = -q(Y,X)$, while metrics are those which satisfy $q(X,Y) = q(Y,X)$ (symmetry vs antisymmetry) and also a condition that $q(X,X) \geq 0$ and is nonzero wherever $X$ is nonzero.

It might be helpful to think about this just as what these look like as bilinear maps on a tangent space. A metric looks like an inner product, whereas a form looks like $q(x,y) = -q(y,x)$ which alternates, and has the (actually equivalent) property that $q(x,x) = 0$. This helps us capture the idea from our initial idea of Riemann integration that if we change the orientation of the integration we should change the sign of the integral.