User döni - MathOverflow

Set of plane curves which intersect a fixed curve with given multiplicity

2013-01-28T22:06:48Z

Let $C \subseteq \mathbb{P}^2$ be an irreducible plane curve of degree $d$ and let $p \in C$. For any integer $n \geq 1$, let $V_n$ be the set of all forms $f$ of degree $d-1$, such that $f$ vanishes in $p$ and such that the intersection multiplicity of $C$ and the curve defined by $f$ is at least $n$. If $p$ is a smooth point of $C$, $V_n$ is a vector space. Is this also the case, if $p$ is a singularity of $C$? In particular, I am interested in the case, where the ground field are the complex numbers.

Characterization of powers of multiaffine polynomials

2013-01-20T16:34:42Z

Let $K$ be an algebraic closed field. A polynomial $h \in K[x_1, \ldots , x_n]$ is called multi-affine, if the degree of $h$ in $x_i$ is at most one for all $i$. Let $e_1, \ldots , e_n$ denote the standard basis of $K^n$. Now let $h \in K[x_1, \ldots , x_n]$ be a multi-affine, homogeneous polynomial and let $f=h^N$ for some $N \in \mathbb{N}$. Cleary, $f$ has the following property: For every $1 \leq i \leq n$ and $v \in K^n$, the univariate polynomial $f(t e_i + v) \in K[t]$ is a power of some univariate polymomial of degree at most one. I conjecture, that the converse is also true: Every homogeneous polynomial with this property is some power of a multi-affine polynomial. It seems to me, that there should be an easy proof or counterexample, but I can't see it at the moment.

Simple diophantic equation

2012-11-30T18:04:08Z

We are looking for all integer solutions for the equation $a^b+1=b^a$. We conjecture that there are only the solutions $(0,b),(1,2),(2,3)$. It is easy to see, that if a is odd and b even, there is only the solution $(1,2)$, but we don't see the general case. We think there should be a simple argument.

Morphism between affine spaces of polynomials by setting equal variables

2012-10-30T10:39:49Z

Let $V=\mathbb{C}[x_1^{1}, \ldots , x_1^{m_1}, \ldots , x_n^{1}, \ldots , x_n^{m_n}]_d$ be the vector space of polynomials of degree $d$. Let $W \subseteq V$ be a Zariski closed subset. Consider the map $\Phi : V \to \mathbb{C}[x_1, \ldots , x_n]_d, f \mapsto f(x_1,\ldots, x_1, \ldots, x_n, \ldots ,x_n)$ obtained by equating $x_i^{1}, \ldots , x_i^{m_i}$ for all $i$. (This map is linear in the coefficients of $f$.) Are there some results about the restriction map $\Phi|_W$? If not in general, perhaps for some non-trivial special cases? Perhaps realized as some quotient map of a group action?

Invariants of group action: SL_n acts simultaneously on m symmetric matrices

2012-09-16T19:20:28Z

Let $\rm{SL}_n$ be the special linear group and let $\rm{Sym}_n$ be the set of all symmetric matrices of size n. $\rm{SL}_n$ acts on $(\rm{Sym}_n)^m$ by $g(A_1, \ldots , A_m)=(gA_1 g^{\rm T}, \ldots , g A_m g^{\rm T})$. Clearly, in the case of $m=1$ the ring of invariants is generated by $\det(A)$. But what are the invariants of this group action in general? Is there an easy description of the ring of invariants, e.g. by giving the generators?

Image of linear projection

2012-09-06T12:14:39Z

Let $X \subset \mathbb{P}^n$ be a projective variety (i.e. zariski closed), and let $\pi : X \dashrightarrow \mathbb{P}^m$ a linear projection ($\pi$ is not in every point of $X$ defined).

Under which conditions is the image $\pi(X)$ zariski closed? What criterions are known?

Group action on Grassmannian: Intersection of two special invariant rings

2012-07-21T16:58:54Z

Let $K$ be a field with characteristic $0$.

Let $G:=G(d,nd)$ the Grassmannian of all $d-$dimensional subspaces of $K^{nd}$ and let $H:=O_d(K)^n$ the n-fold direct product of the orthogonal group. $H$ operates on $G$ in the following way: Let $v_1, \ldots , v_d \in K^{nd}$, $U_1, \ldots , U_n \in O_d(K)$ and let $U \in K^{nd \times nd}$ be the block diagonal matrix with entries $U_1, \ldots , U_n$. We define the group operation $(U_1, \ldots , U_n) ([v_1 \wedge \ldots \wedge v_d]):= [Uv_1 \wedge \ldots Uv_d]$. I am interested in the quotient variety $G//H$. Is there already a description of it?

If not, I considered the group action of $H$ on a open affine $H-$invariant subset of $G$ and reduced it to the following question:

For $1 \leq k \leq n$ and $1 \leq i,j \leq d$ consider the variables $x_{ij}^k$ (thereby $k,i,j$ are indices). Let $R:=K[\sum_{m=1}^d x_{mi}^k x_{mj}^k : 1 \leq k \leq n, 1 \leq i,j \leq d]$ and $S:=K[\sum_{m=1}^d x_{im}^k x_{jm}^l: 1 \leq k,l \leq n, 1\leq i,j \leq d]$. In other words: $R$ is generated by all forms $([x_{1i}^k, \ldots , x_{di}^k] , [x_{1j}^k, \ldots , x_{dj}^k])$, and $S$ is generated by all forms $([x_{i1}^k, \ldots , x_{id}^k] , [x_{j1}^l, \ldots , x_{jd}^l])$. $(\cdot, \cdot)$ denote the standard scalar product. What are the generators of $R \cap S$? ($R \cap S$ is the invariant ring of the described group action on an affine subset)

singular value decomposition

2012-07-06T19:53:32Z

Given regular matrices $A_i,B_i \in \textrm{GL}_n(\mathbb{R}),$ $i=1,2$.

Let $A_1 = U_1 B_1 V_1$ and $A_2=U_2 B_2 V_2$ where $U_i,V_i \in \textrm{O}_n(\mathbb{R})$ $(i=1,2)$ are orthogonal matrices. This means that $A_1$ and $B_1$ resp. $A_2$ and $B_2$ have the same singular values.

Now let $A_2 A_1^{-1}$ and $B_2 B_1^{-1}$ have the same singular values.

I assume that in this case one can find $U_i',V' \in \textrm{O}_n(\mathbb{R})$ so that $A_1 = U_1' B_1 V$ and $A_2=U_2' B_2 V$.

Is this true, at least under some conditions? Or is there a simple counterexample?

Image under "square-morphism"

2012-06-07T19:13:40Z

Let $V \subset \mathbb{P}^n$ be a projective variety. The homogeneous vanishing ideal is generated by the forms $f_1, \ldots , f_r$. Now consider the morphism

$\phi: \mathbb{P}^n \to \mathbb{P}^n, (x_0 : \ldots : x_n) \mapsto (x_0^2: \ldots : x_n^2).$

Is there a way to find generators of the vanishing ideal of the image $\phi(V$) under this morphism, i.e. an easier or more conceptional way than computung the image via Groebner bases?

Comment by Döni

2013-01-29T10:24:37Z

okay thank you, this settles exactly the assumptions I made. Is it at least true, that Vn is always a finite (union) of vector spaces?

Comment by Döni

2013-01-29T10:20:20Z

Thank you, thats interesting. Does this implie, that $V_n$ is always at least some (finite) union of vector spaces? This assumption about degree $d-1$ I only stated to get no common components of $C$ and $f$.

Comment by Döni

2013-01-29T10:16:55Z

thanks, although I wanted $C$ to be irreducible, I think I know where this is going.

Comment by Döni

2013-01-22T17:31:43Z

has been settled.

Comment by Döni

2012-11-30T19:13:57Z

thank you emil, this is an easy solution.

Comment by Döni

2012-11-30T18:41:25Z

okay, thank you.

Comment by Döni

2012-09-16T19:21:41Z

Yes, that is true. Yours is a more natural formulation of my problem. I will pose a new question: mathoverflow.net/questions/107340/…

Comment by Döni

2012-07-07T17:13:00Z

Thank you. This is exactly the kind of result I was looking for.

Comment by Döni

2012-07-07T06:45:58Z

Hello Kjetil, thanks for your comments. Actually the $B_i$ are not supposed to be diagonal. The condition says: $A_i$ and $B_i$ have the same singular values. Perhaps I should rewrite the question. But of course if my assumption is true, it should also hold for diagonal matrices $B_1,B_2$. In your comment, you didn't use the fact, that $A_2 A_1^{-1}$ and $B_2 B_1^{-1}$ have the same singular values. It is clear, that if we drop this condition, the assumption will be false.

Comment by Döni

2012-06-10T15:12:56Z

Okay, thank you. Considering symmetry of the $f_1, \ldots , f_r$ could lead to an acceptable result.