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Insights The Pantheon of Derivatives - Part IV - Comments

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  1. Mar 22, 2017 #1

    fresh_42

    Staff: Mentor

  2. jcsd
  3. Mar 22, 2017 #2
    The definition of the Lie algebra ##g## of a Lie group ##G## is strange (near formula (18)). It looks like the operaton ##[\cdot,\cdot]## defined in ##g## does not have any relation to the group ##G##.

    Actually one of the standard definitions is as follows. Consider two vectors ##a,b\in g##. And let ##L_a:G\to G## be the left shift. Then one can construct vector fields $$A(x)=(dL_x)a,\quad B(x)= (dL_x)b$$ Let ##C(x)## be the commutator of the vector fields ##A(x),B(x)##. Then by definition ##[a,b]=C(e)##, here ##e## is the identical element.

    It would be good to note that equations (40),(41) determine ##\nabla_X## uniquely
     
    Last edited: Mar 22, 2017
  4. Mar 22, 2017 #3

    fresh_42

    Staff: Mentor

    That's right and I removed "on G" as the intention was to define a Lie algebra without a (unnecessary) reference to a group, which already happened prior to the formal definition.
     
  5. Mar 22, 2017 #4
    Formula (42) does not follow from the sentence above. The operator ##\nabla_X## has been previously defined on the vector fields. Thus the left side of this formula is indefinite. Actually this formula is a definition of the operator ##\nabla_X## on tensors of type (2,0). The fact that the manifold is Riemann as well as the dimension of the manifold have not relation to this formula. One can put it by definition for any connection on any manifold.
    Accept the following axiom: $$\nabla _X\langle f,u\rangle=\langle \nabla _X f,u\rangle+\langle f, \nabla _X u\rangle,$$
    for any vector fields ## u,X## and for any covector field ##f##. This axiom
    defines the operator ##\nabla _X## for covector fields and then formula similar to (42) extends the operator ##\nabla _X## to tensor fields of any type ##(p,q)##.

    ps In (42) the symbol ##\times## is tensor product I guess. ##\otimes## is commonly used
     
    Last edited: Mar 22, 2017
  6. Mar 22, 2017 #5

    fresh_42

    Staff: Mentor

    Nope, it's said and meant to be the cross-product.
     
  7. Mar 22, 2017 #6
    then what is cross product of vector fields?

    upd: o, I see! ok
     
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