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Direct sum and direct product

  1. Jul 13, 2014 #1
    The definition (taken from Robert Gilmore's: Lie groups, Lie algebras, and some of their applications):

    We have two vector spaces [itex]V_1[/itex] and [itex]V_2[/itex] with bases [itex]\{e_i\}[/itex] and [itex]\{f_i\}[/itex]. A basis for the direct product space [itex]V_1\otimes V_2[/itex] can be taken as [itex]\{e_i\otimes f_j\}[/itex]. So an element w of this space would look like (summation convention):

    [tex]w=A^{ij}e_i\otimes f_j [/tex]

    For the direct sum space [itex]V_1\oplus V_2[/itex], we take as basis: [itex]\{e_1,e_2,....;f_1,f_2,...\}[/itex].
    \end of stuff from Gilmore

    My question:

    If we take [itex]V_1[/itex] to be the x-axis, and [itex]V_2[/itex] to be the y-axis, we can say that the tensor product space is the y=x line. Since any element would look like: [itex]w=A\;\;\hat{x}\otimes \hat{y} [/itex]


    whereas the direct sum space is spanned by [itex]\{\hat{x},\hat{y}\}[/itex], i.e. it consists of the entire [itex]R^2[/itex].

    but it seems weird to me that the tensor product space in this example has a smaller dimension than the direct sum space. Obviously I have a misconception, where is it?

    Thanks!
     
    Last edited: Jul 13, 2014
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  3. Jul 13, 2014 #2

    micromass

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    I don't know why you call this a direct product. It's the tensor product, it's very different from the direct product. In fact, the direct sum and direct product (of two factors) coincide in this case.

    In what sense is it the y=x line?

    What do the hats mean on ##\hat{x}##?

    Yes, sometimes the tensor product space has a smaller dimension than the direct sum. A lot of times it has a greater dimension however. Why does this seem weird to you? Do you think that a sum should always be smaller than a product? What about ##1\cdot 1<1+1##?
     
  4. Jul 13, 2014 #3
    Some physicists use tensor product and direct product synonymously. The definition given above is the direct product according to Gilmore - he defines the tensor product on groups only - see page 28 of the above mentioned reference.

    In the sense that this space is spanned by [itex]\hat{x}\otimes\hat{y}=(1,1)[/itex] which is a vector in [itex]R^2[/itex] that makes a 45 angle with the x-axis.

    Unit vectors.
    [/QUOTE]
    But the direct (or tensor?) product [itex]R\otimes R[/itex] is simply [itex]R^2[/itex]. So I must be misunderstanding something. Or maybe it's Gilmore? (highly doubtful :))
     
    Last edited: Jul 13, 2014
  5. Jul 14, 2014 #4

    micromass

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    That Gilmore book is awful. He defines the tensor product of vector spaces, not of groups. And he really shouldn't use the term direct product.

    I don't follow at all. Why would ##\hat{x}\otimes \hat{y} = (1,1)##? That makes no sense at all.

    Not at all. We have that ##\mathbb{R}\otimes\mathbb{R}## is the same as ##\mathbb{R}##. A basis for ##\mathbb{R}## is simply ##\{1\}## and a basis for ##\mathbb{R}\otimes\mathbb{R}## is ##\{1\otimes 1\}##.
    In general, if ##V## has dimension ##n## and if ##W## has dimension ##m##, then ##V\otimes W## has dimension ##m\cdot n## and ##V\oplus W## has dimension ##m+n##.
     
  6. Jul 14, 2014 #5
    So what is this item ##\hat{x}\otimes \hat{y}##?

    More importantly, what is the vector space spanned by it? I know it is ##V_1\otimes V_2##, but what is the geometric picture of this space?
     
  7. Jul 14, 2014 #6

    micromass

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    I don't really think there is an easy geometric picture of this.

    For ##\mathbb{R}\otimes\mathbb{R}## this is isomorphic to ##\mathbb{R}## and the isomorphism sends ##x\otimes y## to ##xy##. So in this sense, the tensor product generalizes the usual product on ##\mathbb{R}##.

    So that is one way of seeing the tensor product. It provides a multiplication operation on the vector spaces. But unlike the usual multiplication, the multiplication of a vector in ##V_1## and ##V_2## lies in an entirely new space ##V_1\otimes V_2##. Still, this is a useful intuition, particularly when you go reading about Fock spaces and such.

    The best way of seeing the tensor product is with bilinear maps (or multilinear maps). In that sense, a tensor ##x\otimes y## is in fact related to a bilinear map. For more information, see the free book linear algebra done wrong http://www.math.brown.edu/~treil/papers/LADW/LADW.html Chapter 8.
     
  8. Jul 14, 2014 #7
    Thank you for the clarification, and for the excellent reference.
     
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