Recent content by huyichen

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    Associativity of Tensor product

    It is a well fact that tensor product is associative up to isomorphism, but how should I use Universal property(you know, diagrams that commute) to show that it is true?
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    Periodic and Nonconstant Integral Curve: A Contradiction

    M is a smooth manifolds, and X is a vector field on M, y is a maximal integral curve of X. Now suppose y is periodic and nonconstant, show that there exists a unique positive number T(called the period of y) such that y(t)=y(t') if and only if t-t'=kT for some integer k.(For this problem, What...
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    Maps B2->S2 Define F+ & F-: Show Int w S2 = Int F+*(w) B2 - Int F-*(w)

    But as F- and F+ agrees on equator, then the limit indeed hold, right?
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    Maps B2->S2 Define F+ & F-: Show Int w S2 = Int F+*(w) B2 - Int F-*(w)

    Actually, this is problem 14-6 from Introduction to Smooth Manifolds, Lee's book, we can not directly talk about the boundary because F+, F- is continuous but not smooth up to the equator. And the integrand is unbounded, but if we interpret in an appropriate limiting sense, then we can show...
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    Maps B2->S2 Define F+ & F-: Show Int w S2 = Int F+*(w) B2 - Int F-*(w)

    Justify means that we can indeed replace Integral F^*w with the limit integral as R--> 1
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    Maps B2->S2 Define F+ & F-: Show Int w S2 = Int F+*(w) B2 - Int F-*(w)

    Define F+:B2->S2 by F+=(u,v)=(u,v,sqrt(1-u^2-v^2)) and F-:B2->S2 by F-=(u,v)=(u,v,-sqrt(1-u^2-v^2)) Then to show that Integral of w on S2=Integral of F+^*(w) on B2-Integral of F-^*(w), why do we need to justify the limits(As the integral on the right hand side are defined as limits as...
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    Nonvanishing section for direct sum of Mobius band

    For a direct sum of Mobius band, it is trivial if it has two linear independent nonvanishing sections. I have the following as my sections: s1=(E^(i*theta), (Cos(theta/2), Sin(theta/2)) s2=(E^(i*theta), (-Sin(theta/2), Cos(theta/2)) Clearly, the above sections are linearly independent and...
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    Vector field as smooth embedding

    We can show that any vector field V:M->TM(tangent bundle of M) is smooth embedding of M, but how do we show that these smooth embeddings are all smoothly homotopic? How to construct such a homotopy?
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    How to show that a transverse intersection is clean, but not conversely?

    If K and L are embedded manifold of M, and T_p(K intersect L)=T_p K intersect T_p L and K intersect L is again a embedded manifold , then we say K intersect L is clean
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    How to show that a transverse intersection is clean, but not conversely?

    How to show that a transverse intersection is clean, but not conversely?
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    Number of Nondegenerate Zeros of Vector Field Bounded

    "if a vector field has only nondegenerate zeros then the number of zeros is bounded" With no idea how to show that without using Poincare-Hopf Theorem. Any proof possible without using any concept from algebraic topology?
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    Smooth covering map and smooth embedding

    Now F:S^2->R^4 is a map of the following form: F(x,y)=(x^2-y^2,xy,xz,yz) now using the smooth covering map p:S^2->RP^2, p is the composition of inclusion map i:S^2->R^3 and the quotient map q:R^3\{0}->RP^2. show that F descends to a smooth embedding of RP^2 into R^4. Is the problem asked to...
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    Intersection of complex sphere and cone

    Show the intersection of complex sphere (|z1|^2+|z2|^2+|z3|^2=1) in C^3 and the complex cone (z1^2+z2^2+z3^3=1) in C^3 is a smooth submanifold of C^3. I am trying to do it using regular level set, but I am not sure which one of (1,0) or (1,1,0) should be set to be the regular value?
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    Proving Pushforward Product Isomorphism: M1 x ... x Mk to M1 + ... + Mk

    But, don't we need to take component functions of arbitrary X into account?
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    Smooth manifold and constant map

    Suppose M and N are smooth manifold with M connected, and F:M->N is a smooth map and its pushforward is zero map for each p in M. Show that F is a constant map. I just remember from topology, the only continuous functions from connected space to {0,1} are constant functions. With this be...
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