User peter kronheimer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:18:54Z http://mathoverflow.net/feeds/user/16193 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93871/strong-contactomorphism-group-inside-contactomorphism-group/93975#93975 Answer by Peter Kronheimer for strong contactomorphism group inside contactomorphism group Peter Kronheimer 2012-04-13T16:59:39Z 2012-04-13T16:59:39Z <p>The map isn't surjective on $\pi_{0}$. Take $M$ to be a disjoint union of two 3-spheres, put any contact form $\beta$ on the first sphere, and put $\beta/2$ on the second sphere. There is an element of $\mathrm{Cont}^{+}(M)$ that simply interchanges the two pieces. But this map is not isotopic to a map that preserves the contact form. (Just think of the 3-form $\beta \wedge d\beta$ and the volumes of the two 3-spheres.)</p> http://mathoverflow.net/questions/83523/smooth-four-manifolds-with-contractible-universal-cover/83524#83524 Answer by Peter Kronheimer for Smooth four-manifolds with contractible universal cover Peter Kronheimer 2011-12-15T13:58:48Z 2011-12-15T13:58:48Z <p>In algebraic geometry, there are examples of "fake projective planes", which in this context means smooth complex surfaces of general type with the same cohomology ring as the complex projective plane. It is known that the universal cover of such spaces is the complex hyperbolic ball. So the answer to your question is yes. (The first such fake projetive plane was shown to exist by Mumford.)</p> http://mathoverflow.net/questions/78373/does-the-bertini-theorem-imply-that-there-exists-k-points-such-that-passing-thr/78795#78795 Answer by Peter Kronheimer for Does the Bertini Theorem imply that there exists $k$ points such that passing through them imposes linearly independent conditions? Peter Kronheimer 2011-10-21T21:09:51Z 2011-10-21T21:09:51Z <p>It is quite easy to give $k$ points imposing independent conditions. Just choose the points $p_1,\dots p_k$ one at a time as follows:</p> <p>After choosing the first $j$ points for some <code>$j&lt;k$</code>, take any curve $C$ passing through these $j$ points. Pick any point not lying on $C$. Call this point $p_{j+1}$ and continue. The effect is that the sets of curves passing through the first $j$ points are strictly decreasing sets as $j$ increases, which is all you that you want to achieve.</p> http://mathoverflow.net/questions/78621/kahler-structure-on-a-complex-reductive-group/78670#78670 Answer by Peter Kronheimer for Kähler structure on a complex reductive group Peter Kronheimer 2011-10-20T12:38:33Z 2011-10-20T12:38:33Z <p>Isn't the answer no in the very simplest case? If $K$ is the circle group, then the Kähler structure on the cotangent bundle makes it metrically a cylinder <code>$R \times S^{1}$</code>. I believe this cylinder cannot be isometrically embedded in <code>$C^n$</code> (apply the maximum modulus principal to the derivative of the map).</p> http://mathoverflow.net/questions/75381/relative-version-of-symplectic-thom-conjecture/77697#77697 Answer by Peter Kronheimer for Relative version of Symplectic Thom conjecture. Peter Kronheimer 2011-10-10T14:06:03Z 2011-10-10T14:06:03Z <p>I think this must be a consequence of the version of the slice-Bennequin inequality proved by Mrowka and Rollin (but I might be wrong). Perhaps the argument also requires the boundary to have the "strong filling" property (Stein near the boundary).</p> <p>Given a Legendrian knot $L$ in (to start with) the 3-sphere $S^3$, that inequality says <code>$$2 g_*(L) - 1 \ge tb(L) - r(L).$$</code> Every transverse knot $K$ is a push-off of a Legendrian approximation $L$, and the self-linking number of $K$ is related to the invariants of $L$ by $$sl(K) = tb(L) - r(L).$$ So the slice-Bennequin inequality says <code>$$2 g_*(K) -1 \ge sl(K).$$</code> Unless I'm mistaken, this inequality is an equality for the case of a transverse knot bounding a symplectic surface in the 4-ball.</p> <p>All this generalizes to the case of a symplectic 4-manifold $X$ that is Stein near its (contact) boundary. Given a Legendrian knot $L$ in the boundary, and a homology class $s$ of surfaces in $X$ with boundary $L$, one has invariants $tb(L,s)$ and $r(L,s)$. Then there is an inequality (as in Mrowka-Rollin), <code>$$2 g_*(L,s) - 1 \ge tb(L,s) - r(L,s),$$</code> where <code>$g_*(L,s)$</code> is the smallest possible genus of a surface in the class $s$ with boundary $L$. In terms of a transverse push-off $K$, one again has <code>$$2 g_*(K,s) - 1 \ge sl(K,s).$$</code> If $K$ is actually the transverse boundary of a symplectic surface, then one has equality, this being (I think) just the adjunction formula in a relative version.</p> <p>The Mrowka-Rollin version of the result can today be deduced from the existence of concave fillings (caps). We may assume from the outset that $tb(L,s)$ is positive: if it is not, we may sum in a bunch of Legendrian trefoils until it is. Now enlarge $X$ by first adding a 2-handle along $L$ (standard contact surgery) and then closing it up with a concave filling. The inequality one wants is just the adjunction inequality applied to the homology class formed from $s$ and the 2-disk in the core of the handle.</p> <p>So with less notation, the answer to the original question is supposed to be: take a Legendrian approximation to the transverse knot and alter things so as to make $tb$ postive; then add a 2-handle and a cap to get a closed symplectic manifold. Then apply the adjuntion inequality to the homology class formed from the symplectic surface and the Lagrangian 2-disk.</p> http://mathoverflow.net/questions/83523/smooth-four-manifolds-with-contractible-universal-cover/83524#83524 Comment by Peter Kronheimer Peter Kronheimer 2011-12-15T22:11:03Z 2011-12-15T22:11:03Z I don't think I know any examples with n bigger than 1. I agree that this is an interesting question. http://mathoverflow.net/questions/83523/smooth-four-manifolds-with-contractible-universal-cover/83524#83524 Comment by Peter Kronheimer Peter Kronheimer 2011-12-15T14:45:25Z 2011-12-15T14:45:25Z The previous comment is off-base, because hyperbolic 4-manifolds have signature zero and cannot be definite if their intersection form is non-trivial. http://mathoverflow.net/questions/83523/smooth-four-manifolds-with-contractible-universal-cover/83524#83524 Comment by Peter Kronheimer Peter Kronheimer 2011-12-15T14:20:21Z 2011-12-15T14:20:21Z I don't know of examples where the cohomology ring is that of the connected sum of n projetive planes for n bigger than 1. For small n I think one could look for hyperbolic 4-manifolds, but hyperbolic examples cannot exist for large n. http://mathoverflow.net/questions/83375/connected-sum-decomposition-of-a-knot Comment by Peter Kronheimer Peter Kronheimer 2011-12-14T13:45:47Z 2011-12-14T13:45:47Z A straightforward comment is that this problem is at least as hard as recognizing the unknot. (Given a diagram D for knot, form a larger diagram as the &quot;connected sum&quot; of D with a trefoil diagram. The larger diagram represents a composite knot iff D represents the unknot.) http://mathoverflow.net/questions/78373/does-the-bertini-theorem-imply-that-there-exists-k-points-such-that-passing-thr/78795#78795 Comment by Peter Kronheimer Peter Kronheimer 2011-10-21T21:12:18Z 2011-10-21T21:12:18Z Actually, this is just what Francesco said above. http://mathoverflow.net/questions/75381/relative-version-of-symplectic-thom-conjecture/77697#77697 Comment by Peter Kronheimer Peter Kronheimer 2011-10-10T15:36:06Z 2011-10-10T15:36:06Z Two other thoughts: (1) You probably don't need the &quot;strong&quot; contact condition at the boundary in order to add the contact handle, so just contact-type boundary may be fine here. (2) Another take on the argument might be that one can apply the symplectic Thom conjecture to surfaces that are only semi-symplectic (the form restricted to the surface is non-negative), provided that the homology class has positive self-intersection. This is because you can alter the symplectic form in the tubular neighborhood, to make it positive on the surface.