# Tagged Questions

**-3**

votes

**1**answer

267 views

### Lie algebraic Grassmannian

Assume that $L$ is a Lie algebra structure on $\mathbb{R}^{n}$, and $1<k<n$ is given.
We define $Gr(k,n)_{L}$, the space of all $k$ dimensional Lie subalgebra of $(\mathbb{R}^{n}, L)$.
For ...

**4**

votes

**0**answers

111 views

### formal smooth morphism with a formal smooth source

Let $f:X\rightarrow Y$ a morpism between $k$-schemes ( $k$ a field).
We suppose that X is formally smooth and f is formally smooth and surjective.
Do we have that $Y$ is formally smooth?
Or if it's ...

**6**

votes

**1**answer

303 views

### Is a smooth cubic threefold diffeomorphic to a rational threefold?

A theorem of Clemmens and Griffiths states that a smooth hypesurface in $\mathbb CP^4$ of degree three is not rational. I would like to know if nevertheless it is diffeomorphic (as a smooth real ...

**3**

votes

**2**answers

235 views

### Real analytic submanifolds of $\mathbb{R}^{n}$

Hallo,
Let $(M,g)$ be a Riemannian $k$-dim real analytic submanifold of $\mathbb{R}^{n}$. Is it true that $M$ in $\mathbb{R}^{n}$ looks locally (in a small neigbourhood around some point in $M$) as ...

**9**

votes

**1**answer

476 views

### Is every closed embedded codimension-n submanifold cut out by a section of a rank-n vector bundle?

Let $M$ be a smooth manifold (over $\mathbb R$) and $N \hookrightarrow M$ a closed embedding. Locally near any point in $N$, I can find coordinates $x^1,\dots,x^{\dim M}$ on $M$ so that $N$ is the ...

**6**

votes

**0**answers

254 views

### Universal property for complex blowup in smooth category

If $M$ is a smooth complex manifold and $N$ is a smooth complex surface, we may ask when a holomorphic map $f:M\rightarrow N$ lifts to a map $f:M\rightarrow [N:p]$, where $[N:p]$ denotes the blowup of ...

**3**

votes

**1**answer

407 views

### Fixed points of the action of an algebraic group

Hello!
If a compact Lie group $K$ acts smoothly on a smooth manifold $M$, then the set $M^K$ of fixed points under this action is a smooth submanifold of $M$. This is proved for example in ...

**18**

votes

**2**answers

2k views

### Universal property of the tangent bundle

If $X$ is a scheme (over some base scheme, but which I will ignore) its tangent bundle $T(X)$ is defined as the relative spectrum of the symmetric algebra of its sheaf of differentials. Combining the ...

**0**

votes

**1**answer

318 views

### Why “smooth Gelfand duality” does not involve a topology on the algebras?

The following question naturally originates from this question
and this one.
While the usual $C^{0}$ Gelfand duality involves a topology on the function algebras considered (it relates compact ...

**14**

votes

**4**answers

1k views

### How to classify the algebras C^∞(M)?

This continues my question about smooth Gelfand-duality. In the book
Juan A. Navarro González & Juan B. Sancho de Salas, C∞-Differentiable Spaces, LNM 1824
it is shown that $M \mapsto ...

**5**

votes

**5**answers

697 views

### smooth Gelfand-duality

Assume $M$ is a compact smooth manifold (without boundary). What can we say about the spectrum of the $\mathbb{R}$-algebra $A=C^{\infty}(M)$? The elements of $M$ give rise to rational points of $A$, ...

**5**

votes

**2**answers

405 views

### Question on transversal slice of Lie group

Assume we have action of Lie group $G$ on a manifold $X$. Fix some orbit $\mathcal{O}$, it is known there exist transversal slice $S$ with respect to this orbit. Fix some point $x$ in $\mathcal{O}$, ...

**15**

votes

**1**answer

2k views

### GAGA and Chern classes

My question is as follows.
Do the Chern classes as defined by Grothendieck for smooth projective varieties coincide with the Chern classes as defined with the aid of invariant polynomials and ...