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Quotient space

  1. Jun 19, 2012 #1
    Hello friends,
    I would ask if anyone knows the métrqiue on the quotient space? ie, if one has a metric on a vector space, how can we calculate the metric on the quotient space of E?
  2. jcsd
  3. Jun 19, 2012 #2
    First, what do you mean with quotient space?? Do you mean the quotient space as defined in topology?? In that case, the quotient of a metric space is not in general metrizable.

    If you're working with a normed vector space V and a closed subset W, then V/W (as vector space quotient) does carry a norm which is given by
    [tex]\|x+W\|=inf \{\|x+w\|~\vert~w\in W\}[/tex]
    Last edited: Jun 19, 2012
  4. Jun 19, 2012 #3
    yes I speak of a normed vector space, but how you got this metric? you can give me the link or document where I can find the information?
    and thank you very much.
  5. Jun 19, 2012 #4
  6. Jun 19, 2012 #5
    The elements of the quotient space V/W are each hyperplanes which lie parallel to and have equal dimension to W. Since W is the only one passing through the origin, it is the zero of the space.

    For instance, ℝ2/Y, where Y is the y-axis, is essentially the x-axis. This is because each line parallel to the Y axis (each hyperplane parallel to the space we are dividing by) is totally determined its value measured along the x-axis.

    The formula that micromass gave computes the shortest distance from any point in the plane to any point in the "zero plane." This is the simplest, most canonical way to extend a norm on the original space to a norm on the quotient space.
  7. Jun 20, 2012 #6
    By "closed subset W" you mean linear subspace?
  8. Jun 20, 2012 #7
    I mean a linear subspace that is closed (for the norm), that is: if a sequence in W converges then its limit point is contained in W. We need the subspace to be closed in order for the quotient-norm to define a norm and not a semi-norm.

    And yes, I made a mistake in my post, I needed W to be both closed and a subspace. Just being closed is obviously not enough.
  9. Jun 20, 2012 #8
    I see. I guess this is necessary only for infinite dimensional case, correct?
  10. Jun 20, 2012 #9
    Yes. Finite dimensional subspaces are automatically complete (and thus closed).
  11. Jun 20, 2012 #10
    Thank you for your answers.
  12. Jun 20, 2012 #11
    I have another question now, if we have a vector bundle with a metric E1 G1 and E2 a vector bundle with a metric g2, it will be the metric on the product?
  13. Jun 20, 2012 #12
    "I have another question now, if we have a vector bundle E1 with a metric g1 and E2 a vector bundle with a metric g2, what will be the metric on the product?
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