Let $M$ be a smooth manifold and let $SM$ be the bundle of symmetric bi-linear forms on $TM.$ Riemannian metrics are a particular kind of sections in this bundle. Since any manifold admits a global Riemannian metric it follows that the Euler class of $SM$ is zero. If $M$ is parallelizable then $SM$ is clearly trivial. Are there any manifolds where $SM$ is NOT trivial?
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Yes, lots. You can see this by computing characteristic classes using the splitting principle. Let me write $T$ for the tangent bundle and $S^2 T$ for its symmetric square. If we write $T = L_1 + \dots + L_n$ then
$$S^2 T = \sum_{i \le j} L_i \otimes L_j$$
hence, for example, the first Stiefel-Whitney class is
$$w_1(S^2 T) = \sum_{i \le j} w_1(L_i) + w_1(L_j) = \sum_{i < j} w_1(L_i) + w_1(L_j).$$
In this sum $L_i$ appears $n-1$ times, so
$$w_1(S^2 T) = (n-1) w_1(T)$$
which is equal to $w_1(T)$ if $n$ is even. So it suffices to find a smooth manifold of even dimension with $w_1 \neq 0$. We can take, for example, $\mathbb{RP}^2$.
answered Jan 8, 2016 at 2:31
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$\begingroup$ Thank you Qiaochu. Is there any way we can get an upper bound for the number of linearly independent sections of $S^2T$? Is this upper bound easy to find for the non-parallelizable spheres? $\endgroup$ Commented Jan 8, 2016 at 15:16
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1$\begingroup$ @Mike: we can do this by computing its top Pontryagin and Stiefel-Whitney classes. It doesn't produce a very good bound though, since the dimension of $S^2 T$ is quite a bit larger than the dimension of the manifold. $\endgroup$ Commented Jan 8, 2016 at 15:41