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could any one explain me the following paragraph by a simple example?

"a manifold with boundary is understood to be a smooth (real or complex) manifold with a fixed smooth hypersurface. Two functions on a manifold with boundary are called equivalent if one goes over into the other under a diffeomorphism of the manifold that takes the boundary into itself. On the boundary we consider a distinguished point O. The group of germs of diffeomorphisms of a manifold with boundary at a distinguished point that keep the boundary fixed acts on the spaces of germs and jets of functions at the distinguished point for which this is a critical point with critical value zero"

I know what is manifold with boundary but never saw such a definition or remark as the author said in 1st line, so I am not feeling anything of the first paragrgaph, but I am confident that if any one give example and tell me I can understand.

I know what is critical points like say $f:N\rightarrow M$ be a smoothh map, a point $p\in N$ is said to be a critical point of $f$ if the differential $$f_{*,p}:T_p\rightarrow T_{f(p)}M$$ fails to be surjective and I also know one result for a real valued funtion $f:M\rightarrow \mathbb{R}$, a pt. $p\in M$ is critical iff relative to some chart $(U,x_1,\dots,x_n)$ containing $p$ all the partial derivatives $$\frac{\partial f}{\partial x_i}(p)=0$$

there is also some special kind of group and its action is mentioned here, I could not understand that also. Thank you for help.

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1 Answer 1

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Let's take first the example of the square.

The square can be given a structure of smooth manifold : take a homeomorphism from the square to the circle and the smooth structure of the circle gives back a smooth structure on the square.

But hey, in my mind square have corners ! Ok then we have two possibilities : first one is to say, I'll give an emdedding from 4 points to the square (our smooth manifold) to precise what I consider as corners. With this method you can transform a circle into a triangle, a square etc.

The second method is to say that the circle have charts to $\mathbb R$ and the square have charts to $\mathbb R_+$. In any cases, you choose how you want to describe the topological space.

Then you need to understand what a germ is. Let's take $p \in M$ a point, and $f : M \to \mathbb R$ a smooth function. The germ of $f$ at $p$ is the equivalence class of $f$ under the relation : $f \sim_p g$ if and only if there exists an open set $U$ around $p$ such that $f_{|U} = g_{|U}$.

Example : take an holomorphic map $f$ on $\mathbb C$, then for $z \in \mathbb C$, we can write $f(z) = a_0 + a_1 z + a_2 z^2 + ... + a_n z^n + ...$ which is the power serie of $f$ at zero. Then the germ of $f$ at $0$ can be identified as the sequence $a_0, a_1, ...$ It is important to take some time to see why.

In the case of smooth functions, there are much more classes of functions around a given point (i.e germs) and we don't have a nice description.

Now your text is saying : take $M$ and $N$ two smooth manifolds with an embedding $N \to M$. Take one point $p\in N$. And look at all the germs at $p$ of functions $\phi$ such that $\phi : M \to M$ is a difféomorphism and $\phi_{|N} = Id_N$. As the set of such diffeomorphisms (let's call it $\mathrm{Diff}_N(M)$) is a group (under composition), then the set of germs of elements of $\mathrm{Diff}_N(M)$ at $p$ (let's call it $\mathrm{Diff}_N(M)_p$) will also be a group. It is a good thing to check that.

Finally, it is stated that this group $\mathrm{Diff}_N(M)_p$ acts on a special subset of $\mathcal C^{\infty}(M)_p$ the set of all germs of smooth functions at $p$. It is the subset of germs at $p$ of functions $f : M \to \mathbb R$ such that $f(p) = 0$ and $df_p = 0$. This action is by composition. It is not difficult to do the details.

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