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Today, I came across the notion of generic Riemannian metrics for the first time. Some googling around informed me of the "definition" of what it means for a Riemmanian metric to be generic (see, for example, this post). But I find it hard to verify these conditions. Moreover, my question is "soft", in the sense that I don't require to verify the genericity of a specific metric; I'm just curious about the "existence" of such metrics.

Question: does every (compact?) smooth manifold $M$ (which is assumed to be second-countable, Hausdorff, and connected) admit at least one generic Riemannian metric?

I'm not sure if this question is trivial. I was thinking that perhaps the restriction of any generic Riemannian metric on any smooth submanifold is generic, so one can consider an embedding of a given smooth manifold $M$ into a Euclidean space (so as to realize $M$ as a submanifold of $\mathbb{R}^n$ for large $n$), take a generic metric $g$ on $\mathbb{R}^n$ (one such generic metric certainly exists), and then its restriction $g_{|M}$ to $M$ would be a generic Riemannian metric on $M$.

Is this right? Again, I have not been able to verify the above-indicated conditions for such $g_{|M}$, so I'm not sure and would appreciate some validation (or alternative thoughts). Thanks.

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  • $\begingroup$ What definition of a generic metric are you using here? $\endgroup$ Commented Apr 16 at 14:55
  • $\begingroup$ As mentioned in the original post, the definition of generic is from this post: mathoverflow.net/q/497278 $\endgroup$ Commented Apr 16 at 21:01

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In the MathOverflow post you link to, equations (2.15), (2.16), (2.17) are at each point a linear system of equations for the tensors $H$ and $F$. The vector space of all such tensors $(H,F)$ has dimension $$ R = \frac{n(n+1)}{2}-1+\frac{n(n-1)}{2} = n^2-1.$$ Therefore, the coefficient matrix has $R$ columns. If the number of equations is at least $R$ and the rank of the coefficient matrix is maximal, then the unique solution is $H=F=0$. The set of maximal rank matrices is an open dense subset of the set of all matrices.

Here, there is a nontrivial linear map from the space of Weyl tensors to the space of coefficient matrices. You first want to verify that there exists at least one Weyl tensor that maps to a coefficient matrix with maximal rank $R$. If so, the set of such Weyl tensors is an open dense subset of the space of all Weyl tensors,

If so, then for each $p \in M$, let $\mathcal{R}_p$ be the space of Weyl tensors at $p$ that map to coefficient matrices of rank $R$.

A metric is generic if for each $p\in M$, the Weyl tensor $C_p$ is in $\mathcal{R}_p$. I don't see any straightforward way to prove the existence of a generic metric on a given manifold $M$. My only guess is that if the dimension of $M$ is high enough, you might be able to use convex integration to prove this.

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A covariant derivative $\mathbf{ \mathbb \nabla}$ defined freely by a connection field $x\to \Omega(x)$, acting on vector fields $x\to \mathbf X(x)$, locally spanned by partial derivatives $(x,i) \to (\partial_i f)(x)$ of scalar fumctions $f$ at a point $x$ by

$$\mathbf{ \mathbb \nabla} \ \mathbf X(x) = \partial_{x_k} \mathbf X(x) + (\Omega(\mathbf X))(x) $$ with

$$\left(\ \left(\Omega(\mathbf \partial_i)\right) (x)\ \right)_k= \sum_j \ \Gamma^{j}{}_{\ i,k} \ \partial_j $$

In order $\Gamma$ to represent a Levi-Civita connection $\Omega$, there are two conditions:

  1. Symmetry

    the connection must be torsion free, i.e. transport of the local frame is free of rotations, ie. in any local orthonormal frame for any small 2d-square in any of the planes $ik$, the matrix $\Gamma^j_{ik}$ parallel transports the local frame around its four edges, only allowing metric scalings.

    This condition is trivially checked by symmetry of the lower indices of $\Gamma$. Symmetric connections are called torsion-free

  2. Constance of the two scalar maps, form evaluation and metric product. This is essence the fact that

    $$\nabla (\mathbf d x^i \cdot \partial_k) = \nabla \delta^i_k =0 $$.

    Interpretation of $\delta$ as the mixed form of the metric $g$ implies, that the metric is covariant constant too, ie. the bilinear function $$\_ \cdot \_\ : \quad (x,y)\to \sum_{ik}g^{ik} x_i y_k $$ does not contribute a third term by the product rule in metric contractions. In the same way the Christoffel symbols are derived from linear combinations of cyclic rotations of the gradient the metric $$\partial_i g_{jk}$$ coefficients, the Koszul formula yields the integrability conditions for the index lowered connection coefficients

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    $\begingroup$ This does not answer the question $\endgroup$ Commented Apr 16 at 10:53
  • $\begingroup$ Yes, I know, the open question remains, if every manifold (Hausdorf, compact, second countable, smooth) admits a covarariant derivative with differentiable coefficients. Not my problem. $\endgroup$ Commented Apr 16 at 11:17
  • $\begingroup$ No, the question asks if there always exists a metric satisfying a specific condition (that OP called "generic"). Your answer just says "there are Riemannian metrics" without addressing the asked property $\endgroup$ Commented Apr 16 at 11:19
  • $\begingroup$ Even the refernced post cannot answer the 'generic' question, that is not to be special. Generic metrics are dense, so question in this point is 'yes'. $\endgroup$ Commented Apr 16 at 11:46
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    $\begingroup$ @RolandF just to clarify, are you saying that the answer to the question "does every smooth manifold admit a generic Riemannian metric" is yes? If so, is my proof sketch of obtaining such via restriction of a generic Euclidean metric incorrect? $\endgroup$ Commented Apr 16 at 12:38

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