8
$\begingroup$

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?

$\endgroup$

1 Answer 1

9
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thank you very much. I get it now, so a $2$-form just has the properties that $\omega(X,Y)=-\omega(Y,X)$ (at every point $p\in M$). It doesn't have to be that $\omega(X,Y)$ is smooth. While a $(2,0)$-tensor has the only property of being a smooth function on $M$ for every 2 tangent vectors? $\endgroup$
    – user162520
    Commented Aug 5, 2014 at 4:33
  • $\begingroup$ No, a (2,0) also has a very special property. Letting $X$ be any non-zero tangent vector at $p$, we must have $q(X,X) > 0$, and $q(X,Y) = q(Y,X)$ (note here we have no sign change). The difference can also be seen like this: let $(x_1,x_2) =x, (y_1,y_2) = y$ be two vectors in the plane: a metric is a bilinear map that looks a lot like $(x,y) \mapsto x \cdot y$ the Euclidean inner product, whereas a 2-form looks a lot like $$\det\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}$$ Also, it doesn't make sense to say $q(X,Y)$ is smooth for tangent vectors $X,Y$ at a point p... $\endgroup$
    – jxnh
    Commented Aug 5, 2014 at 4:37
  • $\begingroup$ I meant that if $X,Y$ are two smooth vector fields (so they give a tangent vector at each point, then $q(X,Y)$ gives a smooth function. $\endgroup$
    – jxnh
    Commented Aug 5, 2014 at 4:38
  • $\begingroup$ Ok thank you. I meant that $q(X,Y)$ is a smooth function of $M$ for any two vector fields, not tangent vectors. Ok so a $(2,0)$ tensor is symmetric and satisfies $q(X,X)>0$ and a $2$-form satisfies just antisymmetry. $\endgroup$
    – user162520
    Commented Aug 5, 2014 at 4:41
  • 1
    $\begingroup$ @user162520: No, a metric tensor is a (2,0)-tensor which also has the properties of being symmetric and positive definite. $\endgroup$ Commented Aug 5, 2014 at 6:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .