When in differential geometry one shows , on a riemannian manifold, that a (unique) connection exists, (Levi Civita connection), is it possible to "lift" that notion to the principal bundle of frames over that manifold?In that case, is that lifted connection unique? (if so, the Yang Mills field would precisely be the LV connection)

Does the bundle of frames always admit a one form, even if the manifold has no metric on it?In that case that would provide a lifting to the LC connection...

Quoting from this blog of Prof Terry

“…as the very useful Sobolev embedding theorem, which allows one to trade regularity for integrability…”

Thats one use of Sobolev spaces. In this context, Fourier and Plancheral methods come very handy, when the corresponding Sobolev space is also a Hilbert space. But that is not always the case…. Only $L^2$ based Sobolev spaces are Hilbert spaces. According to Sobolev emebdding, if the function in $\mathbb{R}^d$ need to be holder continuous, then its gardient needs to be $L^p$ integrable with $p >= d+1$. So for $d >1$, we need $p > 2$, so the associated Sobolev space cannot have a Hilbert space structure. So in this context, we cannot use Fourier Plancheral techniques. “But if” (stress If)… I say, that I can always find a Hilbert space, for any d, (even for cases when d>1), how useful a tool that it would be, in the context of PDE. What would the impact be? Any examples of PDE, on which there would be impact? Appreciate your valuable comment.

The quotient manifold theorem says that

If $G$ is a Lie group acting freely and properly on a smooth manifold $M$ then $M/G$ has a (unique) smooth structure such that the projection $\pi:M\to M/G$ is a submersion.

I was wondering what happens when the action is not free. My intuition suggests that we get corners, I have in mind this example: The action of $\frac{\mathbb{Z}}{2\mathbb{Z}}$ over $\mathbb{S}^2$ induced by the reflection wrt the $zy$-plane. The quotient manifold obtained is $\mathbb{D}^2$ and the boundary $\partial\mathbb{D}^2 $ can be identified with the fixed points of the action i.e. $\mathbb{S}^2\cap zy \text{-plane}$.

Does anyone know a theorem that covers the non-free case? Where can I read about it?

Suppose $M$ is a smooth connected complete Riemannian manifold of dimension $n\geq 2$. Let $d:M\times M\rightarrow \mathbb{R}^+$ be the distance induced by the Riemannian metric on $M$. For $p\in M$ we set $d_p:=d(p,\cdot)$. We know that $d_p$ is smooth on $M\setminus (C_p\cup\{p\})$, where $C_p$ is the cut locus of $p$, which is a null set according to the Riemannian measure on $M$. Moreover, $d_p$ is regular in any point $q\in M\setminus (C_p\cup\{p\})$, since its gradient at $q$ is the derivative of the unique minimal geodesic at $d_p(q)$ joining $p$ and $q$.

For $R>0$ consider the level set $d_p^{-1}(R)$. Since $R$ does not need to be a regular value of $d_p$, we may not be able to define a normal vector field globally on $d_p^{-1}(R)$. Is there some characterisation of the intersection $C_p\cap d_p^{-1}(R)$, which states that set is "small" maybe in the sense of some $N-2$-dimensional Hausdorff measure or in some topological sense? Or is there some other way to define a unit normal vector field "almost everywhere" on $d_p^{-1}(R)$?

Assume that $f$ is smooth function defined in the unit disk $D: x^2+y^2\le 1$, and consider the integral $$I=\int_D f dxdy=\int_0^1r \int_0^{2\pi} f(re^{it})dt.$$

Then it is clear that for $r\in[0,1]$ there is $t_r\in [0,2\pi]$ so that $$I=2\pi \int_0^1 r f(re^{it_r})dr.$$ My question is, can we choose $t_r$ to depend smoothly on $r$.

Can you solve this problem for me?

*Premise*

Let $K$ be a field of characteristic zero and $f\in K[X_1,\dots,X_m]$. By Hironaka's theorem, there exists a log resolution (over $K$) of the ideal $(f)$. Let $\{(N_i,\nu_i)\}_i$ be the numerical data of a fixed log resolution. The quantity $$ lct_K(f):=\min_{i}\frac{\nu_i}{N_i} $$ does not depend on the choice of the log resolution and it is called the log canonical threshold of $f$ over $K$.

*Questions*

Let $f\in \mathbb{Q}[X_1,\dots,X_m]$. By definition, we have $$ lct_{\mathbb{Q}}(f)\ge lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) \ge lct_{\mathbb{C}}(f_{\mathbb{C}}). $$ On the other hand, from Denef's formula for the motivic Igusa zeta function it follows that for all but finitely many $p$ one has $$ lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) \ge lct_{\mathbb{Q}}(f_{\mathbb{Q}}). $$ This shows that $$ lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) = lct_{\mathbb{Q}}(f_{\mathbb{Q}}) \quad \forall\forall p. $$

**1. Is this equality actually true for all $p$?**

In all the counterexamples I have found in the literature for the validity of Denef's formula for the "bad" primes (in the sense of Denef) one still has $lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) = lct_{\mathbb{Q}}(f_{\mathbb{Q}})$ also for bad primes $p$. Were this not always the case, has anybody a counterexample at hand?

**2. What can we say about the comparison with $lct_{\mathbb{C}}(f_{\mathbb{C}})$?**