What is an example of a non-Riemannian manifold?

1.8k Views Asked by At

A Riemannian manifold $(M,g)$ is a manifold equipped with a metric tensor ($g$) that is both symmetric and positive definite. Now if the metric tensor is symmetric but not positive definite, then it's a pseudo-Riemannian manifold. But what the case of a manifold equipped with a metric tensor $(M,g)$ such that $g$ is not symmetric(regardless of whether it is or is not positive definite)?

That is, $\forall (x,y) \ such \ that (\ x \neq y \ ) \in T_{p}(M) \ g(x,y)\neq g(y,x)$ at any point p in M.

1

There are 1 best solutions below

0
On BEST ANSWER

Let me first try to make sense of the question (in its current revised version; my guess is that this is not the last revision). Recall that a metric tensor by the very definition is required to be symmetric and positive definite. Hence, a "nonsymmetric metric tensor" is an oxymoron. My best guess is that OP really means not a metric tensor, but a covariant 2-tensor $g$ on a manifold $M$. The nonsymmetry condition, as stated in the revised version, says:

For every $p\in M$ and every two distinct vectors $x, y$ in $T_pM$, we have $g(x,y)\ne g(y,x)$.

Note that if $\lambda\in {\mathbb R}$ is different from $1$ and $x\in T_pM$ is a nonzero vector, then (by the definition of a tensor!) $$ g( \lambda x, x)= \lambda g(x, x)= g(x, \lambda x). $$ Hence, for $y=\lambda x$, we get $g(x,y)=g(y,x)$. Thus, there are no tensors satisfying the required inequality.

OK, maybe OP really means that $x, y$ are linearly independent in the stated assumption (nonvanishing of $x, y$ is then automatic). Then the nonsymmetry requirement becomes: $$ g(x, y)\ne g(y, x) $$ for every $p\in M$ and all linearly independent vectors $x, y\in T_p$. Then tensors with this property do exist in dimension 2. For instance, one can take the area form $\omega=dx_1\wedge dx_2$ on the plane $M$. Then for all $p\in M, x, y\in T_pM$ we have $$ \omega(x,y)=-\omega(y,x)\ne \omega(y,x) $$ as long as $x, y$ are linearly independent.

It is a nice linear algebra exercise to see that tensors satisfying my nonsymmetry condition do not exist on manifolds of dimension $\ge 3$. In particular, symplectic forms (on manifolds of dimension $\ge 4$) fail this nonsymmetry condition.