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I Trouble with exact sequence

  1. Sep 15, 2016 #1
    I am trying to up my understanding on Manifolds, Tensors and Forms by reading a book of that title by Paul Renteln.

    I have got stuck fairly early on by his use of "exact sequences". Can some one give me a concrete example of a shot exact sequence of the form 0 -> U->V->W->0 where U,V,W are vector spaces 0 is the null vector space and the -> are linear maps not necessarily the same one each time. I do think I understand the ker of U and im U etc.

    I have tried various web pages but they give abstract examples which don't seem to help and I am obviously in a gumption trap!

    I trust this is the right place to ask and this is personal study out of interest and in no way a homework assignment!

    Regards Andrew
     
  2. jcsd
  3. Sep 15, 2016 #2

    jedishrfu

    Staff: Mentor

  4. Sep 15, 2016 #3
    Thanks, I did look at them but they were too abstract given my inexperience in this field (pun intended).

    Regards Andrew
     
  5. Sep 15, 2016 #4

    lavinia

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    Map the plane considered as a vector space of the real numbers onto the real line also considered as a vector space over the real numbers by projecting onto the x coordinate. This is a linear mapping with kernel the y-axis. That is: the kernel is the subspace of the plane of vectors whose x coordinate is zero.

    One has the exact sequence of vector spaces and linear maps,

    0→y-axis →plane→ real line →0

    where the first two arrows are inclusion maps and the second two are projections.

    The inclusion of the y-axis into the plane is injective so the sequence is exact at the y-axis. The projection of the plane onto the real line is surjective with kernel the y-axis. So the sequence is exact at the plane and at the real line.
     
    Last edited: Sep 15, 2016
  6. Sep 15, 2016 #5

    fresh_42

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    Another "common" example is ##\{0\} \rightarrowtail 2 \cdot \mathbb{Z} \rightarrowtail \mathbb{Z} \twoheadrightarrow \mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}_2 \twoheadrightarrow \{0\}## or any other number than ##2##.
     
  7. Sep 15, 2016 #6

    lavinia

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    The idea of an exact sequence applies not only to vector spaces but to algebraic objects in general.

    For instance @fresh_42 gave an exact sequence of abelian groups in post #5. If one considers an abelian group to be a module of the ring of integers then one has an exact sequence of modules over a ring. This idea extends to modules over arbitrary rings.

    The general idea is that the arrows must be homomorphisms of the objects in the sequence,
     
    Last edited: Sep 15, 2016
  8. Sep 15, 2016 #7
    Thanks these examples should me get going again. Much appreciated.
    Regards Andrew
     
  9. Sep 15, 2016 #8

    fresh_42

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    Regard the fact in @lavinia's example, that one cannot go back to the y-axis. One has to chose another axis, here the x-axis. Otherwise it won't be exact anymore as the kernel of the first projection wouldn't be the image of the last injection anymore.

    May I ask you about the context? It could be theoretical, geometric, algebraic, as a covering of groups and so on. Even derivations fit in here.

    Edit: My bet: Forms.
     
    Last edited: Sep 15, 2016
  10. Sep 15, 2016 #9

    lavinia

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    The examples given here are usually called "short exact sequences" because they only have three spaces(excluding the end zeros) and four arrows. But exact sequences can have any number of arrows and spaces even infinitely many. One only requires that the image of one arrow is the kernel of the following arrow. Infinitely long exact sequences are common in Algebraic Topology.
     
  11. Sep 15, 2016 #10
    Indeed you may. It was in the context of developing linear algebra of vector spaces. The author of the book I am reading asserts "Exact sequences of vector spaces show up everywhere and satisfy some particularly nice properties so it it is worth exploring them a bit."

    I am trying to improve my knowledge of a number of areas of mathematics to be able to understand the latest developments in physics (relative to me learning physics 40 years ago). However, at my age it take a while to sink in and even more to recall it!

    Regards Andrew
     
  12. Sep 15, 2016 #11

    fresh_42

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    I understand. I'm in a similar situation. I know of some (probably unexplored) long exact sequences and look out what they could be good for.
    And current physics is really far away from what I once had left ...
     
  13. Sep 15, 2016 #12
    A diagram of linear maps between vector spaces

    {0} → V → W → X → {0}​

    is exact (such a sequence is called a "short exact sequence") precisely when these conditions hold:

    1) The map V → W is one-to-one. (I.e., the only vector in V taken to 0 ∈ W is the 0 ∈ V);

    2) The map W → X is onto (i.e., every vector of X is the image of some vector of W);

    3) The image of the map V → W is the kernel of the map W → X.


    This situation is equivalent to finding a subspace V of a vector space W and letting the map

    V → W

    be the inclusion, and the map

    W → X

    be the quotient map

    W → W/V.
     
  14. Sep 17, 2016 #13

    mathwonk

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    a sequence ...-->A-->B-->C-->D-->E-->..... is exact at a point, for example at C, if and only if the image of the incoming map (B-->C) equals the kernel of the outgoing map (C-->D) at that point, and a sequence is exact iff it is exact at all points.

    thus the short sequence 0-->A-->B-->C-->0 is exact at A iff the image of the incoming map 0-->A, which must be jiust the subspace {0}, is the kernel of the outgoing map A-->B, i.e. iff the outgoing map has kernel {0} so is injective.

    Then it is also exact at B iff the image of the incoming injection A-->B, equals the kernel of the outgoing map B-->C. And it is also exact at C iff the kernel of the outgoing map C-->0, which must be all of C, equals the image of the incoming map B-->C, i.e. iff the incoming map B-->C is surjective.

    In general, if ...-->A-->B-->C-->D-->E-->. is any exact sequence, the sequence 0-->(im(A-->B)) --> B --> C --> C/(im(B-->C)) -- > 0 is also exact.

    I.e. a sequence of form 0-->ker(B-->C) --> B -->C --> coker(B-->C) --> 0 is always exact, where map ker(B-->C) --> B is inclusion, and the map C --> coker(B-->C) is the quotient projection. Hence "short" exact sequences of form 0-->A-->B-->C-->D-->0, i.e. with 4 rather than only three intermediate spaces, are very common.

    A sequence of maps such that the image of each incoming map is only contained in the kernel of the outgoing map, is called a "complex". Then one defines the cohomology of a complex at each point, as the quotient of the kernel of the outgoing map by the image of the incoming map, a measure of failure of exactness. Then there is a basic result that to every short exact sequence of complexes 0-->A-->B-->C-->0, there is an associated long exact sequence of cohomology 0-->H^0(A)-->H^0(B)-->H^0(C)-->H^1(A)-->H^1(B)-->H^1(C)-->H^2(A)-->....

    The fundamental exercise is to define the "connecting homomorphism" H^n(C)-->H^(n+1)(A). You might try this exercise, (hint: pull back and push down.)
     
    Last edited: Sep 17, 2016
  15. Sep 17, 2016 #14

    mathwonk

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    sorry for the abstraction. here is the basic example:

    0-->R^n --> R^n x R^m --> R^m -->0. the first map takes a vector v in R^n to the pair (v,0) in R^n x R^m, and the second takes the pair (v,w) in R^n x R^m to the vector w in R^m.

    If f:V-->W is any linear map, and if ker(f) is the subspace of V consisting of those vectors v in V such that f(v) = 0, and if im(f) is the subspace of W consisting of all vectors in W which occur as values of f, then 0-->ker(f) --> V --> im(f) --> 0 is exact.

    and so is 0-->ker(f) -->V -->W --> W/im(f) -->0. with the "obvious" maps. here W/im(f) is called the cokernel of f, so the sequence 0-->ker(f)-->V-->W-->coker(f)-->0 is always exact, and conversely if
    0-->A-->V-->W-->B-->0 is exact then A ≈ ker(V-->W) and B ≈ coker(V-->W).
     
    Last edited: Sep 18, 2016
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