Vector Field Exponential MapReasoning about Lie theory and the Exponential MapLie Group Multiplication in CoordinatesQuestion about SO(3) and “infinitesimal generators”Why is the Lie derivative linear in the vector field?Does the exponential map respect module actions?Induced Lie Algebra Representation, Left invariant vector fields and more…Lie algebra of projective linear group and one-parameter transformationsSecond-order term of Baker-Campbell-Hausdorff formulaExponential of the product between $x$ the derivative operator of $x$ acting in a $f(x)$Can the formula for the derivative of the exponential map (in a Lie group) be derived without using the power series formula?

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A Trivial Diagnosis



Vector Field Exponential Map


Reasoning about Lie theory and the Exponential MapLie Group Multiplication in CoordinatesQuestion about SO(3) and “infinitesimal generators”Why is the Lie derivative linear in the vector field?Does the exponential map respect module actions?Induced Lie Algebra Representation, Left invariant vector fields and more…Lie algebra of projective linear group and one-parameter transformationsSecond-order term of Baker-Campbell-Hausdorff formulaExponential of the product between $x$ the derivative operator of $x$ acting in a $f(x)$Can the formula for the derivative of the exponential map (in a Lie group) be derived without using the power series formula?













1












$begingroup$


I've got $bf v = x^2partial_x$, and I'm trying to find $exp(varepsilonbf v)$, but I'm having some trouble.



If I define $bf v^n+1 = bf vbf v^n$ then I get a different outcome to $bf v^n+1 = bf v^nbf v$.



For example:



$$bf v^2 = bf vv = (x^2partial_x)(x^2partial_x) = x^2(partial_xx^2)partial_x =
x^2(2x)partial_x = 2x^3partial_x$$



$$bf v^3 = bf vbf v^2 =(x^2partial_x)(2x^3partial_x)=x^2(partial_x2x^3)partial_x = x^2(6x^2)partial_x = 6x^4partial_x$$
$$bf v^3 = bf v^2bf v = (2x^3partial_x)(x^2partial_x) = 2x^3(partial_xx^2)partial_x = 2x^3(2x)partial_x = 4x^4partial_x$$



Using $bf v^n+1 = bf vbf v^n$ gives
$$exp(varepsilon bf v)x = fracx1-varepsilon x$$



While using $bf v^n+1 = bf v^nbf v$ gives
$$exp(varepsilon bf v)x = fracx2(1+mathrm e^2varepsilon x)$$



In both cases, when $varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$lim_varepsilon to 0 fracmathrm dmathrm dvarepsilon exp(varepsilon bf v)x = bf v$$



The same $bf v in mathfrak g$ can't possible generate two different flows, can it?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
    $endgroup$
    – Eric Towers
    Mar 14 at 17:37










  • $begingroup$
    If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
    $endgroup$
    – Fly by Night
    Mar 14 at 18:16











  • $begingroup$
    And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
    $endgroup$
    – Eric Towers
    Mar 15 at 14:03
















1












$begingroup$


I've got $bf v = x^2partial_x$, and I'm trying to find $exp(varepsilonbf v)$, but I'm having some trouble.



If I define $bf v^n+1 = bf vbf v^n$ then I get a different outcome to $bf v^n+1 = bf v^nbf v$.



For example:



$$bf v^2 = bf vv = (x^2partial_x)(x^2partial_x) = x^2(partial_xx^2)partial_x =
x^2(2x)partial_x = 2x^3partial_x$$



$$bf v^3 = bf vbf v^2 =(x^2partial_x)(2x^3partial_x)=x^2(partial_x2x^3)partial_x = x^2(6x^2)partial_x = 6x^4partial_x$$
$$bf v^3 = bf v^2bf v = (2x^3partial_x)(x^2partial_x) = 2x^3(partial_xx^2)partial_x = 2x^3(2x)partial_x = 4x^4partial_x$$



Using $bf v^n+1 = bf vbf v^n$ gives
$$exp(varepsilon bf v)x = fracx1-varepsilon x$$



While using $bf v^n+1 = bf v^nbf v$ gives
$$exp(varepsilon bf v)x = fracx2(1+mathrm e^2varepsilon x)$$



In both cases, when $varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$lim_varepsilon to 0 fracmathrm dmathrm dvarepsilon exp(varepsilon bf v)x = bf v$$



The same $bf v in mathfrak g$ can't possible generate two different flows, can it?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
    $endgroup$
    – Eric Towers
    Mar 14 at 17:37










  • $begingroup$
    If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
    $endgroup$
    – Fly by Night
    Mar 14 at 18:16











  • $begingroup$
    And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
    $endgroup$
    – Eric Towers
    Mar 15 at 14:03














1












1








1





$begingroup$


I've got $bf v = x^2partial_x$, and I'm trying to find $exp(varepsilonbf v)$, but I'm having some trouble.



If I define $bf v^n+1 = bf vbf v^n$ then I get a different outcome to $bf v^n+1 = bf v^nbf v$.



For example:



$$bf v^2 = bf vv = (x^2partial_x)(x^2partial_x) = x^2(partial_xx^2)partial_x =
x^2(2x)partial_x = 2x^3partial_x$$



$$bf v^3 = bf vbf v^2 =(x^2partial_x)(2x^3partial_x)=x^2(partial_x2x^3)partial_x = x^2(6x^2)partial_x = 6x^4partial_x$$
$$bf v^3 = bf v^2bf v = (2x^3partial_x)(x^2partial_x) = 2x^3(partial_xx^2)partial_x = 2x^3(2x)partial_x = 4x^4partial_x$$



Using $bf v^n+1 = bf vbf v^n$ gives
$$exp(varepsilon bf v)x = fracx1-varepsilon x$$



While using $bf v^n+1 = bf v^nbf v$ gives
$$exp(varepsilon bf v)x = fracx2(1+mathrm e^2varepsilon x)$$



In both cases, when $varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$lim_varepsilon to 0 fracmathrm dmathrm dvarepsilon exp(varepsilon bf v)x = bf v$$



The same $bf v in mathfrak g$ can't possible generate two different flows, can it?










share|cite|improve this question









$endgroup$




I've got $bf v = x^2partial_x$, and I'm trying to find $exp(varepsilonbf v)$, but I'm having some trouble.



If I define $bf v^n+1 = bf vbf v^n$ then I get a different outcome to $bf v^n+1 = bf v^nbf v$.



For example:



$$bf v^2 = bf vv = (x^2partial_x)(x^2partial_x) = x^2(partial_xx^2)partial_x =
x^2(2x)partial_x = 2x^3partial_x$$



$$bf v^3 = bf vbf v^2 =(x^2partial_x)(2x^3partial_x)=x^2(partial_x2x^3)partial_x = x^2(6x^2)partial_x = 6x^4partial_x$$
$$bf v^3 = bf v^2bf v = (2x^3partial_x)(x^2partial_x) = 2x^3(partial_xx^2)partial_x = 2x^3(2x)partial_x = 4x^4partial_x$$



Using $bf v^n+1 = bf vbf v^n$ gives
$$exp(varepsilon bf v)x = fracx1-varepsilon x$$



While using $bf v^n+1 = bf v^nbf v$ gives
$$exp(varepsilon bf v)x = fracx2(1+mathrm e^2varepsilon x)$$



In both cases, when $varepsilon =0$, we get just $x$, i.e. the identity element. Also, in both cases, we get
$$lim_varepsilon to 0 fracmathrm dmathrm dvarepsilon exp(varepsilon bf v)x = bf v$$



The same $bf v in mathfrak g$ can't possible generate two different flows, can it?







lie-groups lie-algebras vector-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 14 at 17:20









Fly by NightFly by Night

26.1k32978




26.1k32978











  • $begingroup$
    Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
    $endgroup$
    – Eric Towers
    Mar 14 at 17:37










  • $begingroup$
    If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
    $endgroup$
    – Fly by Night
    Mar 14 at 18:16











  • $begingroup$
    And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
    $endgroup$
    – Eric Towers
    Mar 15 at 14:03

















  • $begingroup$
    Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
    $endgroup$
    – Eric Towers
    Mar 14 at 17:37










  • $begingroup$
    If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
    $endgroup$
    – Fly by Night
    Mar 14 at 18:16











  • $begingroup$
    And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
    $endgroup$
    – Eric Towers
    Mar 15 at 14:03
















$begingroup$
Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
$endgroup$
– Eric Towers
Mar 14 at 17:37




$begingroup$
Doesn't it seem odd to you that you are not getting iterates (i.e. nested compositions) of $partial_x$? For instance, if $mathbfv = partial_x$, what should $mathbfv^2$ be?
$endgroup$
– Eric Towers
Mar 14 at 17:37












$begingroup$
If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
$endgroup$
– Fly by Night
Mar 14 at 18:16





$begingroup$
If $bf v = partial_x$ then I would say $bf v^2 = (1partial_x)(1partial_x) = 1(partial_x 1)partial_x=0$. Similarly $bf v^n =0$ for all $n ge 2$, meaning that $exp(varepsilon bf v)x= (1+varepsilon partial_x)x=x+varepsilon$, which is the correct flow.
$endgroup$
– Fly by Night
Mar 14 at 18:16













$begingroup$
And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
$endgroup$
– Eric Towers
Mar 15 at 14:03





$begingroup$
And yet, "the derivative with respect to $x$ of the derivative with respect to $x$" is not the zero operator (without projection to first order operators). If $mathbfv = partial_x$, $mathbfv^2 = partial_x^2$, the second derivative with respect to $x$. (... which projects onto the zero vector on the subspace of first order differentials.) But in short, in your expression "$1(partial_x 1) partial_x$", you have made an error: the first $partial_x$ acts on the rest of the expression; you should have "$1 partial_x (1 partial_x) = 1 (1 partial_x^2 + 0 partial_x)$".
$endgroup$
– Eric Towers
Mar 15 at 14:03











3 Answers
3






active

oldest

votes


















1












$begingroup$

If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
&= x^2 f'(x) text, and \
mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
&= (x^2 partial_x)left( x^2 f'(x) right) \
&= x^2(2x f'(x) + x^2 f''(x) ) text, or \
mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
&= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
&= x^2(2x f'(x) + x^2 f''(x) )
endalign*

so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.



(Of course, if we're linearizing, we project onto the first term in both of those.)






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
    $endgroup$
    – Fly by Night
    Mar 14 at 18:22











  • $begingroup$
    @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
    $endgroup$
    – Eric Towers
    Mar 15 at 13:57











  • $begingroup$
    But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
    $endgroup$
    – Fly by Night
    Mar 16 at 18:22


















1












$begingroup$

You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.



v generates a shift operator, and it pays to define suitable canonical coordinates,
$$
y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
$$

so that you are shifting y by $epsilon$,
$$
e^epsilon partial_y ~~f(y)= f(y+epsilon),
$$

which reads
$$
e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
$$

a standard formula in the RG advection of QFT.



Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.






share|cite|improve this answer











$endgroup$




















    0












    $begingroup$

    The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because $bf v(f)$ is a function and so $bf v[bf v(f)]$ is also a function, while $[bf v(f)]bf v$ is a function times a vector field, i.e. a vector field.



    The differential operator $bf v = x^2partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.



    Applied to $x$, we get $x mapsto 1 mapsto x^2$, meaning $bf v(x) = x^2$.



    Applied to $x^2$, we get $x^2 mapsto 2x mapsto 2x^3$, meaning $bf v^2(x) = 2x^3$.



    Applied to $2x^3$, we get $2x^3 mapsto 6x^2 mapsto 6x^4$, meaning $bf v^3(x) = 6x^4$.



    In general, $bf v^n(x) = n!x^n+1$, and so for all $|varepsilon x|<1$
    begineqnarray*
    sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
    &=& xsum_n ge 0 (varepsilon x)^n \ \
    &=& fracx1-varepsilon x
    endeqnarray*






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      3 Answers
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      3 Answers
      3






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      active

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      active

      oldest

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      1












      $begingroup$

      If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
      mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
      &= x^2 f'(x) text, and \
      mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
      &= (x^2 partial_x)left( x^2 f'(x) right) \
      &= x^2(2x f'(x) + x^2 f''(x) ) text, or \
      mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
      &= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
      &= x^2(2x f'(x) + x^2 f''(x) )
      endalign*

      so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.



      (Of course, if we're linearizing, we project onto the first term in both of those.)






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
        $endgroup$
        – Fly by Night
        Mar 14 at 18:22











      • $begingroup$
        @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
        $endgroup$
        – Eric Towers
        Mar 15 at 13:57











      • $begingroup$
        But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
        $endgroup$
        – Fly by Night
        Mar 16 at 18:22















      1












      $begingroup$

      If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
      mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
      &= x^2 f'(x) text, and \
      mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
      &= (x^2 partial_x)left( x^2 f'(x) right) \
      &= x^2(2x f'(x) + x^2 f''(x) ) text, or \
      mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
      &= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
      &= x^2(2x f'(x) + x^2 f''(x) )
      endalign*

      so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.



      (Of course, if we're linearizing, we project onto the first term in both of those.)






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
        $endgroup$
        – Fly by Night
        Mar 14 at 18:22











      • $begingroup$
        @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
        $endgroup$
        – Eric Towers
        Mar 15 at 13:57











      • $begingroup$
        But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
        $endgroup$
        – Fly by Night
        Mar 16 at 18:22













      1












      1








      1





      $begingroup$

      If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
      mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
      &= x^2 f'(x) text, and \
      mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
      &= (x^2 partial_x)left( x^2 f'(x) right) \
      &= x^2(2x f'(x) + x^2 f''(x) ) text, or \
      mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
      &= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
      &= x^2(2x f'(x) + x^2 f''(x) )
      endalign*

      so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.



      (Of course, if we're linearizing, we project onto the first term in both of those.)






      share|cite|improve this answer









      $endgroup$



      If you apply the product rule carefully, you don't get two different answers. In your example, beginalign*
      mathbfv(f(x)) &= (x^2 partial_x)(f(x)) \
      &= x^2 f'(x) text, and \
      mathbfv^2(f(x)) &= (x^2 partial_x)left( (x^2 partial_x)(f(x)) right) \
      &= (x^2 partial_x)left( x^2 f'(x) right) \
      &= x^2(2x f'(x) + x^2 f''(x) ) text, or \
      mathbfv^2(f(x)) &= left((x^2 partial_x) (x^2 partial_x) right) (f(x)) \
      &= (x^2(2x partial_x + x^2 partial_x^2))(f(x)) \
      &= x^2(2x f'(x) + x^2 f''(x) )
      endalign*

      so we see $mathbfv^2 = 2x^3 partial_x + x^4 partial_x^2$. Similarly, $mathbfv^3 = 6 x^4 partial_x + 6 x^5 partial_x^2 + x^6 partial_x^3$.



      (Of course, if we're linearizing, we project onto the first term in both of those.)







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Mar 14 at 17:56









      Eric TowersEric Towers

      32.9k22370




      32.9k22370











      • $begingroup$
        Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
        $endgroup$
        – Fly by Night
        Mar 14 at 18:22











      • $begingroup$
        @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
        $endgroup$
        – Eric Towers
        Mar 15 at 13:57











      • $begingroup$
        But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
        $endgroup$
        – Fly by Night
        Mar 16 at 18:22
















      • $begingroup$
        Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
        $endgroup$
        – Fly by Night
        Mar 14 at 18:22











      • $begingroup$
        @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
        $endgroup$
        – Eric Towers
        Mar 15 at 13:57











      • $begingroup$
        But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
        $endgroup$
        – Fly by Night
        Mar 16 at 18:22















      $begingroup$
      Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
      $endgroup$
      – Fly by Night
      Mar 14 at 18:22





      $begingroup$
      Thanks for the reply. Am I right to think that $bf v$ is best understood by its action on a function, i.e. $bf v(f)$, and so "multiplying on the right" doesn't make sense. $bf v[bf v(f)]$ is fine, while $[bf v(f)]bf v$ doesn't make sense because it's not a function. Hence $bf v^n+1 = bf vbf v^n$ is the only way to go?
      $endgroup$
      – Fly by Night
      Mar 14 at 18:22













      $begingroup$
      @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
      $endgroup$
      – Eric Towers
      Mar 15 at 13:57





      $begingroup$
      @FlybyNight : $[mathbfv(f)]mathbfv$ is an operator as much as $x^2 partial_x$ is since $mathbfv(f)$ is some function and the $mathbfv$ on the right still has an open slot to consume a function. Note that I showed both $mathbfv(mathbfv(f))$ and $(mathbfv(mathbfv))(f)$ above, so both "wrap another $mathbfv$ on the left" ($mathbfv (mathbfv^n(f))$) and "let $mathbfv^n$ act on $mathbfv(f)$" ($mathbfv^n (mathbfv(f))$), arriving at the same result both ways.
      $endgroup$
      – Eric Towers
      Mar 15 at 13:57













      $begingroup$
      But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
      $endgroup$
      – Fly by Night
      Mar 16 at 18:22




      $begingroup$
      But $bf v(f)bf v$ isn't a function, while $bf vv(f)$ is. That's what I meant.
      $endgroup$
      – Fly by Night
      Mar 16 at 18:22











      1












      $begingroup$

      You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.



      v generates a shift operator, and it pays to define suitable canonical coordinates,
      $$
      y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
      $$

      so that you are shifting y by $epsilon$,
      $$
      e^epsilon partial_y ~~f(y)= f(y+epsilon),
      $$

      which reads
      $$
      e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
      $$

      a standard formula in the RG advection of QFT.



      Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.






      share|cite|improve this answer











      $endgroup$

















        1












        $begingroup$

        You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.



        v generates a shift operator, and it pays to define suitable canonical coordinates,
        $$
        y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
        $$

        so that you are shifting y by $epsilon$,
        $$
        e^epsilon partial_y ~~f(y)= f(y+epsilon),
        $$

        which reads
        $$
        e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
        $$

        a standard formula in the RG advection of QFT.



        Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.






        share|cite|improve this answer











        $endgroup$















          1












          1








          1





          $begingroup$

          You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.



          v generates a shift operator, and it pays to define suitable canonical coordinates,
          $$
          y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
          $$

          so that you are shifting y by $epsilon$,
          $$
          e^epsilon partial_y ~~f(y)= f(y+epsilon),
          $$

          which reads
          $$
          e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
          $$

          a standard formula in the RG advection of QFT.



          Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.






          share|cite|improve this answer











          $endgroup$



          You are looking at the simplest Lie advective flow in perturbation theory (as applied in physics: QFT) and the workhorse example in the 19th century book of Georg Sheffer cited.



          v generates a shift operator, and it pays to define suitable canonical coordinates,
          $$
          y=-1/x, qquad Longrightarrow qquad x^2 partial_x=partial_y ,
          $$

          so that you are shifting y by $epsilon$,
          $$
          e^epsilon partial_y ~~f(y)= f(y+epsilon),
          $$

          which reads
          $$
          e^epsilon x^2partial_x ~~g(x)= gleft(frac-1y+epsilonright )=gleft (fracx1-epsilon xright ),
          $$

          a standard formula in the RG advection of QFT.



          Your difficulties are traceable to your refusal to perform your Heaviside calculus manipulations of functions of differential operators with a "test-function" f(x) on the right, which keeps track of the proper chain rule action of noncommuting derivative operators.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 16 at 2:51

























          answered Mar 15 at 16:59









          Cosmas ZachosCosmas Zachos

          1,810522




          1,810522





















              0












              $begingroup$

              The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because $bf v(f)$ is a function and so $bf v[bf v(f)]$ is also a function, while $[bf v(f)]bf v$ is a function times a vector field, i.e. a vector field.



              The differential operator $bf v = x^2partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.



              Applied to $x$, we get $x mapsto 1 mapsto x^2$, meaning $bf v(x) = x^2$.



              Applied to $x^2$, we get $x^2 mapsto 2x mapsto 2x^3$, meaning $bf v^2(x) = 2x^3$.



              Applied to $2x^3$, we get $2x^3 mapsto 6x^2 mapsto 6x^4$, meaning $bf v^3(x) = 6x^4$.



              In general, $bf v^n(x) = n!x^n+1$, and so for all $|varepsilon x|<1$
              begineqnarray*
              sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
              &=& xsum_n ge 0 (varepsilon x)^n \ \
              &=& fracx1-varepsilon x
              endeqnarray*






              share|cite|improve this answer











              $endgroup$

















                0












                $begingroup$

                The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because $bf v(f)$ is a function and so $bf v[bf v(f)]$ is also a function, while $[bf v(f)]bf v$ is a function times a vector field, i.e. a vector field.



                The differential operator $bf v = x^2partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.



                Applied to $x$, we get $x mapsto 1 mapsto x^2$, meaning $bf v(x) = x^2$.



                Applied to $x^2$, we get $x^2 mapsto 2x mapsto 2x^3$, meaning $bf v^2(x) = 2x^3$.



                Applied to $2x^3$, we get $2x^3 mapsto 6x^2 mapsto 6x^4$, meaning $bf v^3(x) = 6x^4$.



                In general, $bf v^n(x) = n!x^n+1$, and so for all $|varepsilon x|<1$
                begineqnarray*
                sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
                &=& xsum_n ge 0 (varepsilon x)^n \ \
                &=& fracx1-varepsilon x
                endeqnarray*






                share|cite|improve this answer











                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because $bf v(f)$ is a function and so $bf v[bf v(f)]$ is also a function, while $[bf v(f)]bf v$ is a function times a vector field, i.e. a vector field.



                  The differential operator $bf v = x^2partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.



                  Applied to $x$, we get $x mapsto 1 mapsto x^2$, meaning $bf v(x) = x^2$.



                  Applied to $x^2$, we get $x^2 mapsto 2x mapsto 2x^3$, meaning $bf v^2(x) = 2x^3$.



                  Applied to $2x^3$, we get $2x^3 mapsto 6x^2 mapsto 6x^4$, meaning $bf v^3(x) = 6x^4$.



                  In general, $bf v^n(x) = n!x^n+1$, and so for all $|varepsilon x|<1$
                  begineqnarray*
                  sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
                  &=& xsum_n ge 0 (varepsilon x)^n \ \
                  &=& fracx1-varepsilon x
                  endeqnarray*






                  share|cite|improve this answer











                  $endgroup$



                  The action of a vector field as a derivation on the space of functions means that it only makes sense to multiply on the left because $bf v(f)$ is a function and so $bf v[bf v(f)]$ is also a function, while $[bf v(f)]bf v$ is a function times a vector field, i.e. a vector field.



                  The differential operator $bf v = x^2partial_x$ is best understood as differentiate with respect to $x$, and then multiply by $x^2$.



                  Applied to $x$, we get $x mapsto 1 mapsto x^2$, meaning $bf v(x) = x^2$.



                  Applied to $x^2$, we get $x^2 mapsto 2x mapsto 2x^3$, meaning $bf v^2(x) = 2x^3$.



                  Applied to $2x^3$, we get $2x^3 mapsto 6x^2 mapsto 6x^4$, meaning $bf v^3(x) = 6x^4$.



                  In general, $bf v^n(x) = n!x^n+1$, and so for all $|varepsilon x|<1$
                  begineqnarray*
                  sum_n ge 0 fracvarepsilon^nn!bf v^n(x) &=& sum_n ge 0 fracvarepsilon^nn!n!x^n+1 \ \
                  &=& xsum_n ge 0 (varepsilon x)^n \ \
                  &=& fracx1-varepsilon x
                  endeqnarray*







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 17 at 1:40

























                  answered Mar 16 at 19:05









                  Fly by NightFly by Night

                  26.1k32978




                  26.1k32978



























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