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Derivations vs Directional derivatives

  1. Mar 14, 2014 #1
    In some books, when discussing the relation between partial/directional derivatives and tangent vectors, one makes a generalization called a "derivation". A derivation at ##\vec{a} \in \mathbb{R}^n## is defined as a linear map ##D: C^{\infty}(\mathbb{R}^n) \to \mathbb{R}## which for ##f,g \in C^\infty(\mathbb{R}^n)## that satisfies

    ##D(fg) = f(a) Dg + g(a) Df.##

    On the other hand, some books just stick to directional derivatives, so I wondered: what is the virtue of introducing derivations?

    Can someone give an example of something that is a derivation, but not a directional derivative?
     
  2. jcsd
  3. Mar 21, 2014 #2
    Derivations are a more general term. One definition of the tangent space at a point ##p## on a manifold ##M## is that the tangent space ##T_pM## is the vector space of all derivations at ##p##. However, I can't think of anywhere where you can't think of elements of the tangent space as directional derivatives.

    In Lie algebras, the map ##\operatorname{ad}_x:\mathfrak{g}\to\mathfrak{g}## that takes ##y\in\mathfrak{g}## to ##[x,y]\in\mathfrak{g}## (The ##[\cdot,\cdot]## is called the bracket. It's the multiplication operation on ##\mathfrak{g}##.) is a derivation, but it isn't a directional derivative.
     
  4. Mar 22, 2014 #3

    Fredrik

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    If v is a derivation at p, and ##x:U\mapsto\mathbb R^n## is a coordinate system such that ##p\in U##, then there exist real numbers ##v^1,\dots,v^n## such that
    $$v=\sum_{k=1}^n v^k\frac{\partial}{\partial x^k}\!\bigg|_p.$$ If this right-hand side is what you consider a directional derivative, then this means that every derivation is a directional derivative.

    Not sure if ##\mathrm{ad}_x## should really be considered a derivation, since it's ##\mathfrak g##-valued rather than ##\mathbb R##-valued.
     
  5. Mar 22, 2014 #4

    micromass

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    It's still called a derivation: http://en.wikipedia.org/wiki/Derivation_(differential_algebra [Broken])

    Anyway, the Lie bracket can be seen as an actual derivative if you use the notion of Lie derivatives, so its not too farfetched to call it a derivation.
     
    Last edited by a moderator: May 6, 2017
  6. Mar 22, 2014 #5
    (...ERMAGERD Micro's back!!! :biggrin:)

    Micro is correct. There is a construction in differential geometry known as the Lie derivative, which can be applied to various tensor fields. We define the Lie derivative of a vector field ##Y## with ##X## to be ##\mathcal{L}_X(Y)=[X,Y]##.

    To generalize, for any given noncommutative algebra ##A##, the map ##D_x: y\mapsto xy-yx## is a derivation.

    Proof: ##D_x(yz)=xyz-yzx=xyz-yxz+yxz-yzx=D_x(y)z+yD_x(z)##.
     
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