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 Feb24 revised Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem deleted 9 characters in body Feb24 answered Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem Feb24 comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem Also, the statement that $F$ has constant rank (as a smooth map between manifolds) means that the derivative $dF$ has constant rank as a map between tangent bundles. Hence, to fit it in to your context, $dF$ would need to have constant rank as a map of the tangent bundles over the TOTAL spaces $E$ and $E^{\prime}$. Feb24 comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem I don't think you'll be able to use this. Since the total spaces of each bundle is a smooth manifold, certainly $F^{-1}(s(p))$ is an embedded submanifold ($s$ is the zero section, $p$ a point on the base). In fact, it's a subspace of the fiber. The problem is stitching these sub manifolds together smoothly. Feb24 comment Kernel of a Constant Rank Bundle Map Using the Constant Rank Theorem What do you mean by the kernel of a smooth map. If you mean it's derivative, then this would be a special case of the claim where $E$ is the tangent bundle. Feb9 revised Question on tangent spaces misspelled question Feb9 suggested approved edit on Question on tangent spaces Feb8 answered What to answer when people ask what I do in mathematics Jan30 comment Algebraic multiplicity of an eigen value Algebraic multiplicity is just the degree of the linear term corresponding to $\lambda$ in the characteristic polynomial. Every matrix is similar to an upper triangular matrix and the determinant is invariant under similarity. Is this not enough to prove the claim? Jan12 comment Homomorphism between homotopy groups of spheres induced by the fibration and the multiplication map of $SO(n)$ @archipelago: Did you read my answer? I gave a large class of examples where $\lambda=0$. Jan12 answered Homomorphism between homotopy groups of spheres induced by the fibration and the multiplication map of $SO(n)$ Jan11 comment Conditions for the integral to equal zero The integral is nonzero even when $f(x)=1$. If your asking if $f(x)$ has to be bounded when the integral is finite, this is also not true. Consider, $f(x)=x$. Jan11 comment Nonlinear eigenvalue problem $Ax = f(c) x$ I'm a bit confused about the set up. If the $v_{i}'s$ are eigenectors with eigenvalues $b_{i}$, are they not precisely the eigenvectors for $f(c_{i})=b_{i}$. Is the base field real or complex? Is $A$ nice in any way? (e.g. diagonalizable, hermetian, unitary, orthogonal, real self-adjoint) Jan11 comment $k$-jets , submanifolds I think your observation is the key step. $J^{1}(n,1)$ should just be the dual of $\mathbb{R}^{n}$, $\Sigma^{n-1}$ is all nonzero functionals and $\Sigma^{n}$ contains just the $0$ functional. Are these submanifolds? For the second part, perhaps you could use this characterization of $\Sigma^{n}$ along with the definition of a Morse function. Jan11 comment What does $d\zeta_1\wedge\cdots\wedge d\zeta_n$ mean in the context of Cauchy formula (on polydiscs)? @TedShifrin: You are absolutely right. I, admittedly, did not read the problem carefully and assumed (a dangerous practice for mathematicians) that the resulting manifold would be complex. Jan10 answered Examples of a Subbasis and Basis of a topology. Jan10 comment What does $d\zeta_1\wedge\cdots\wedge d\zeta_n$ mean in the context of Cauchy formula (on polydiscs)? I'm not a great person to ask for a reference, as complex geometry is a bit outside my field of expertise. However, a brief account can be found in a book by M. Nakahara called "Geometry, Topology and physics. I think it's in Chapter 7, or somewhere around there. Jan10 comment What does $d\zeta_1\wedge\cdots\wedge d\zeta_n$ mean in the context of Cauchy formula (on polydiscs)? Also, I don't think you want the wedge in the denominator. Jan10 comment What does $d\zeta_1\wedge\cdots\wedge d\zeta_n$ mean in the context of Cauchy formula (on polydiscs)? It is the wedge product of forms. You can view $\partial_{0}P$ as a complex manifold. The $n$ fold wedge will give a measure a measure on this manifold which you can integrate against. Dec5 awarded Yearling