# Tag Info

29

The equation $$\frac{x^k-1}{x-1}=y^m$$ is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $m\geq 2$, the only solutions are $$\frac{3^5-1}{3-1}=11^2,\qquad \frac{7^4-1}{7-1}=20^2,\qquad \frac{18^3-1}{18-1}=7^3.$$ For $m=2$, the equation was solved by Ljunggren (Norsk. Mat. Tidsskr. 25 (1943), 17-20). ...

20

Let me expand my comment in a short answer. One of the most common way to build a rational map $f \colon X \dashrightarrow \mathbb{P}(a_1, \ldots, a_n)$ is to consider a line bundle $\mathscr{L}$ on $X$ together with sections $$\sigma_1 \in H^0(X, \, \mathscr{L}^{a_1}), \quad \sigma_2 \in H^0(X, \, \mathscr{L}^{a_2}), \ldots, \sigma_n \in H^0(X, \, \... 20 Here's an easy, direct definition of E_8. The compact Lie group E_8 is the colimit in the category of topological groups of the following diagram of groups$$ {\scriptstyle\begin{matrix} &SU(2)&\\[-1mm] &\downarrow&\\[-1mm] &SU(3)&\\[-1mm] &\uparrow&\\[-1mm] SU(2)\to SU(3) \leftarrow SU(2)&\!\!\!\!\!\to SU(3) \...

20

It suffices to consider the case when $\Omega$ is a circumcircle, so let it be. At first, the points $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic if and only if $$\frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot \frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(\heartsuit)$$ (by Pascal theorem, they lie on a ...

19

This is a nice lemma: I know a good deal of similar results but this one is unknown to me. I believe it is suitable, as an answer, to give a proof that works with no restriction on the cardinality of the underlying field $F$. I will frame the answer in terms of matrix spaces. Thus, we have a linear subspace $V \subset M_{n,p}(F)$ such that, for every non-...

18

The algebraic group $E_8$ is the group of automorphisms of the $E_8$ lattice vertex algebra, by Frenkel-Kac and Segal. This vertex algebra has a self-dual integral form, so the construction works over arbitrary commutative rings. To construct the vertex algebra, one only needs the rank 8 unimodular $E_8$ lattice, not the 248-dimensional $E_8$ Lie algebra.

18

This configuration has automorphisms by the symmetric group $S_5$, and can be identified with the planes $a_i = a_j$ ($0 \leq i < j \leq 4$) in the projective 3-space $a_0+a_1+a_2+a_3+a_4 = 0$, by setting $$(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).$$ the ten planes then have $(i,j)$ in the order $$(0,1),\,(2,3),\,(2,4),\,(0,... 18 The secant variety Sec_k(V^n_2) is the variety parametrizing (n+1)\times (n+1) symmetric matrices modulo scalar of rank at most k that is of corank at least n+1-k. Then by Proposition 12(b) in J. Harris; L. W. Tu, On symmetric and skew-symmetric determinantal varieties, Topology 23 (1984), no. 1, 71–84. the degree of Sec_k(V^n_2) is given by$$\deg(...

17

That is certainly not true. Consider the case that $C$ is an elliptic curve. Then $\text{Aut}(C\times C)$ contains $\text{GL}(2,\mathbb{Z})$ as a subgroup.

15

Perhaps the easiest is to blow up a plane along a fat point: Blow up $\mathbb A^2_k=\mathrm{Spec} k[x,y]$ at the ideal $(x^2,y^2)$. You should get a pinch point (Whitney's umbrella). This itself is an interesting singularity. It's simple normal crossing away from the pinch point, but not so simple there. If you want to practice blow-ups, then as a second ...

15

From four subspaces in general position one can generate an infinite number of other subspaces by closing up under joins and meets. This is true even for subspaces of $\mathbb{R}^3$ (any field of characteristic zero in place of $\mathbb{R}$ would do). This is easy to see in the corresponding projective plane picture: take $a = (0: 0 : 1)$, $b = (1: 0: 1)$, $... 15 Regarding your question about weighted projective spaces, a lot is known about them, see for instance  and . In particular, any weighted projective space$\mathbb{P}(\mathcal Q)$is irreducible, normal, Cohen-Macaulay and has at most cyclic quotient singularities (hence rational singularities), see , p. 122. However, weighted projective spaces ... 15 It turns out that condition (T) is, indeed, sufficient for the$27$lines (distinct and intersecting as expected) to lie on a cubic surface. To see this, consider the lines$a_1,a_2,a_3,a_4,a_5$and$b_6$, where the labeling is as in note (2) of the question:$a_1$through$a_5$are pairwise skew, and$b_6$intersects all of them. Choose$4$distinct ... 15 There's a straightforward abstract answer that you may not like, but, because it clarifies your question and explains a uniform way to answer similar questions, I'll sketch it here. First, consider a simpler problem of this kind: Suppose that one wants to describe the group of isometries of a Riemannian metric$\rho$on a Riemannian$n$-manifold$M$. By ... 14 Note that without any assumption on the singularities the answer to your question is yes. For instance, consider the hypersurface$Y = \{x_0^2+x_1^3+x_2^4\}\subset\mathbb{P}^3$. Then$Sing(Y) = [0:0:0:1]$. The projective tangent cone of$Y$in the singular point is not reduced. Therefore, the singularity is not ordinary. Let$\pi:X\rightarrow\mathbb{P}^3$be ... 14 Probably final revision: I am indebted to Dave Witte-Morris, who added a reference to a refinement of Zsigmondy's Theorem by W. Feit, of which I was unaware, and pointed out that consequently, a complete answer to the question implicitly followed from what was previously written. In fact, going beyond Dave Witte-Morris's original suggestion (but still ... 13 Actually, though this may seem pedantic, there are two almost-complex structures on$\mathbb{CP}^m$that are invariant under$\mathrm{SU}(m{+}1)$, namely the 'standard' one and its conjugate. Of course, they are equivalent, but only by using an outer automorphism of$\mathrm{SU}(m{+}1)$. The reason this is of more than pedantic interest is that, when you ... 13 Yes, there are other Sylvester-Gallai configurations in$\mathbb{P}^2(\mathbb{C})$. Apart from the Hesse configuration (that contains$9$points) the minimum number of points for a non-collinear configuration is$12$. A configuration with$12$points actually exists over any field$\mathbb{K}$of characteristic different from$2$and containing a square ... 13 Recently Lusztig gave a much simpler definition of$E_8$(and all the simple Lie algebras/groups) that avoids the usual sign issue with the standard Chevalley/Serre construction. See Lusztig - On conjugacy classes in the group$E_8$and Geck - On the construction of semisimple Lie algebras and Chevalley groups. I think this construction of Lusztig's is not ... 13 With the aid of the answer I now managed to find a better realization. At the expense of some of the symmetries it can be nicely drawn in 3d space - it is just the barycentric subdivision of a tetrahedron: The planes are faces (4 of them) and the planes through an edge and the barycenter of the tetrahedron (+6); the lines are edges (6), the lines joining a ... 13 This cannot be done. Let$G_1$and$G_2$be the groups $$G_1 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^4 = (bc)^4 = (ac)^2 \rangle$$ $$G_2 = \langle a,b,c | a^2 = b^2 = c^2 = (ab)^3 = (bc)^3 = (ac)^3 \rangle.$$ You are looking for surjections from the$G_j$onto$PGL_3(\mathbb{F}_2)$, which is the simple group of order$168$. Now,$G_1$and$G_2$are Coxeter ... 13 This is, indeed, true. To prove this, assume we have an embedding$\mathbb{P}^2 \to \operatorname{LGr}(V)$(where$V$is a symplectic vector space). Let$U \subset V \otimes \mathcal{O}$be the tautological subbundle on$\operatorname{LGr}(V)$. Note that it extends to an exact sequence $$0 \to U \to V \otimes \mathcal{O} \to U^\vee \to 0$$ on$\...

12

Reverse math usually means work in subsystems of arithmetic. That goes a bit beyond analysis---Simpson's book has plenty of classic results from the theory of countable groups, rings, and fields, and much of the more recent focus in the area has been countable combinatorics. But all of these are basically about sets of natural numbers and things those can ...

12

In the notes on chapters at the end of Geometric Stability Theory, I give references. For Fact 1.11 it is Doyen and Hubaut, Finite regular locally projective geometries, Math. Zeitschrift, 1971. Zilber's book, Uncountably categorical theories, Translations of Math. monographs, vol 117, AMS, 1993, also mentions this on p. 27.

12

Here is a less algebraic and more topological answer: it's known that any compact (Hausdorff) topological ring must be totally disconnected. In particular, there's no hope for either $\mathbb{RP}^1$ or $\mathbb{CP}^1$ to have the structure of a topological ring. (As Ben Webster mentions, $\mathbb{CP}^1$ also can't even be a topological group.)

12

EDIT: I've just realized that this holds under somewhat weaker assumptions. It is not necessary that the fibers of $g$ are connected. EDIT#2: Apparently, in my previous edit I weakened the conditions too far... properness of $g$ is back as it is needed, but connectivity of fibers is not as it is not. I think it is OK now. :) Cool question. Actually, I ...

12

For $\mathbb{R}^n$: the fundamental theorem of projective geometry (proof: https://www3.nd.edu/~andyp/notes/FunThmProjGeom.pdf) says that the bijections of $\mathbb{R}^n$ taking lines to lines are the affine maps $x\mapsto Ax+b$ for an invertible matrix $A$ and a constant vector $b$. For $S^n$: a theorem by the same name shows that the bijections of ...

12

Commutative algebra is NOT the same as algebraic geometry, especially projective algebraic geometry. The variety in $\mathbb{P}^9$ defined by $I$ and the variety in $\mathbb{P}^9$ defined by $I_0$ are the same variety. If you were to work in $\mathbb{A}^{10}$, then $I$ and $I_0$ would define different affine schemes; the scheme defined by $I$ has some extra ...

11

As pointed out by Alex in his comment, this is in general not true. For instance, consider the case $N=d=n=m=2$. Then $|L|$ is the linear system of plane curves of degree $2$ passing through $p_1$ and $p_2$ with multiplicity $2$. Of course there is only one such a curve, namely the line through $p_1$ and $p_2$ counted with multiplicity $2$. In particular, $|... 11 Not exactly. The quadratic transformation commutes with the action of$S_3$, and they both act on$(\mathbb{C}^*)^2$; so the automorphism group is$(\mathbb{C}^{*})^{2}\rtimes (S_3\times S_2)\$. You'll find a detailed study of the automorphisms of del Pezzo surfaces in Chapter 8 of Dolgachev's book "Classical Algebraic Geometry: a modern view".

Only top voted, non community-wiki answers of a minimum length are eligible