User d&#246;ni - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:18:20Z http://mathoverflow.net/feeds/user/24296 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120151/set-of-plane-curves-which-intersect-a-fixed-curve-with-given-multiplicity Set of plane curves which intersect a fixed curve with given multiplicity Döni 2013-01-28T22:06:48Z 2013-01-29T09:47:04Z <p>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.</p> http://mathoverflow.net/questions/119413/characterization-of-powers-of-multiaffine-polynomials Characterization of powers of multiaffine polynomials Döni 2013-01-20T16:34:42Z 2013-01-20T16:34:42Z <p>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.</p> http://mathoverflow.net/questions/115010/simple-diophantic-equation Simple diophantic equation Döni 2012-11-30T18:04:08Z 2012-11-30T18:04:08Z <p>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.</p> http://mathoverflow.net/questions/111068/morphism-between-affine-spaces-of-polynomials-by-setting-equal-variables Morphism between affine spaces of polynomials by setting equal variables Döni 2012-10-30T10:39:49Z 2012-10-30T10:39:49Z <p>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?</p> http://mathoverflow.net/questions/107340/invariants-of-group-action-sl-n-acts-simultaneously-on-m-symmetric-matrices Invariants of group action: SL_n acts simultaneously on m symmetric matrices Döni 2012-09-16T19:20:28Z 2012-09-17T14:18:58Z <p>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? </p> http://mathoverflow.net/questions/106504/image-of-linear-projection Image of linear projection Döni 2012-09-06T12:14:39Z 2012-09-06T12:14:39Z <p>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).</p> <p>Under which conditions is the image $\pi(X)$ zariski closed? What criterions are known?</p> http://mathoverflow.net/questions/102831/group-action-on-grassmannian-intersection-of-two-special-invariant-rings Group action on Grassmannian: Intersection of two special invariant rings Döni 2012-07-21T16:58:54Z 2012-07-21T16:58:54Z <p>Let $K$ be a field with characteristic $0$.</p> <p>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?</p> <p>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:</p> <p>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)</p> http://mathoverflow.net/questions/101527/singular-value-decomposition singular value decomposition Döni 2012-07-06T19:53:32Z 2012-07-07T09:40:13Z <p>Given regular matrices $A_i,B_i \in \textrm{GL}_n(\mathbb{R}),$ $i=1,2$.</p> <p>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.</p> <p>Now let $A_2 A_1^{-1}$ and $B_2 B_1^{-1}$ have the same singular values.</p> <p>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 $A_2=U_2' B_2 V$.</p> <p>Is this true, at least under some conditions? Or is there a simple counterexample?</p> http://mathoverflow.net/questions/99052/image-under-square-morphism Image under "square-morphism" Döni 2012-06-07T19:13:40Z 2012-06-07T19:13:40Z <p>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 </p> <p>$\phi: \mathbb{P}^n \to \mathbb{P}^n, (x_0 : \ldots : x_n) \mapsto (x_0^2: \ldots : x_n^2).$ </p> <p>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?</p> http://mathoverflow.net/questions/120151/set-of-plane-curves-which-intersect-a-fixed-curve-with-given-multiplicity/120159#120159 Comment by Döni Döni 2013-01-29T10:24:37Z 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? http://mathoverflow.net/questions/120151/set-of-plane-curves-which-intersect-a-fixed-curve-with-given-multiplicity/120189#120189 Comment by Döni Döni 2013-01-29T10:20:20Z 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 $f$. http://mathoverflow.net/questions/120151/set-of-plane-curves-which-intersect-a-fixed-curve-with-given-multiplicity/120158#120158 Comment by Döni Döni 2013-01-29T10:16:55Z 2013-01-29T10:16:55Z thanks, although I wanted $C$ to be irreducible, I think I know where this is going. http://mathoverflow.net/questions/119413/characterization-of-powers-of-multiaffine-polynomials Comment by Döni Döni 2013-01-22T17:31:43Z 2013-01-22T17:31:43Z has been settled. http://mathoverflow.net/questions/115010/simple-diophantic-equation Comment by Döni Döni 2012-11-30T19:13:57Z 2012-11-30T19:13:57Z thank you emil, this is an easy solution. http://mathoverflow.net/questions/115010/simple-diophantic-equation Comment by Döni Döni 2012-11-30T18:41:25Z 2012-11-30T18:41:25Z okay, thank you. http://mathoverflow.net/questions/102831/group-action-on-grassmannian-intersection-of-two-special-invariant-rings Comment by Döni Döni 2012-09-16T19:21:41Z 2012-09-16T19:21:41Z Yes, that is true. Yours is a more natural formulation of my problem. I will pose a new question: <a href="http://mathoverflow.net/questions/107340/invariants-of-group-action-sl-n-acts-simultaneously-on-m-symmetric-matrices" rel="nofollow" title="invariants of group action sl n acts simultaneously on m symmetric matrices">mathoverflow.net/questions/107340/&hellip;</a> http://mathoverflow.net/questions/101527/singular-value-decomposition/101565#101565 Comment by Döni Döni 2012-07-07T17:13:00Z 2012-07-07T17:13:00Z Thank you. This is exactly the kind of result I was looking for. http://mathoverflow.net/questions/101527/singular-value-decomposition/101549#101549 Comment by Döni Döni 2012-07-07T06:45:58Z 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. http://mathoverflow.net/questions/99052/image-under-square-morphism Comment by Döni Döni 2012-06-10T15:12:56Z 2012-06-10T15:12:56Z Okay, thank you. Considering symmetry of the $f_1, \ldots , f_r$ could lead to an acceptable result.