Coordinate-free Hessian

  • #1
ergospherical
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In local coordinates, the hessian of the function ##f## at point ##p## is ##H = \partial_i \partial_k f dx^i \otimes dx^k##. A coordinate-free generalisation is (see) ##H = \nabla df##, or explicitly ##H = \nabla_i (df)_k dx^i \otimes dx^k = \nabla_i \partial_k f dx^i \otimes dx^k##. How is this motivated?
 

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  • #2
martinbn
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In local coordinates, the hessian of the function ##f## at point ##p## is ##H = \partial_i \partial_k f dx^i \otimes dx^k##. A coordinate-free generalisation is (see) ##H = \nabla df##, or explicitly ##H = \nabla_i (df)_k dx^i \otimes dx^k = \nabla_i \partial_k f dx^i \otimes dx^k##. How is this motivated?
How is what motivated?
 
  • #3
ergospherical
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I'm curious to know why ##H = \nabla df## is the correct generalisation of the usual expression (for local coordinates).
 
  • #4
jbergman
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I am not sure if this qualifies but from Lee's Introduction to Riemannian Manifolds, if we have a covariant derivative operator, ##\nabla##, then ##\nabla f## is just the 1-form ##df## because both have the same action on tangent vectors.

$$ \nabla f(X) = \nabla_X f = Xf = df(X) $$

The 2- tensor ##\nabla^2 f## is called the covariant Hessian of f and

$$\nabla^2 f = \nabla(df)$$

The last two formulas are just local coordinate formulas for the above which can be computed from the standard formulas for the covariant derivative.

Lastly, it's action on two tangent vectors is given by,
$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = Y(Xf)-(\nabla_Y X)f$$
 
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  • #5
jbergman
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Lastly, it's action on two tangent vectors is given by,
$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = Y(Xf)-(\nabla_Y X)f$$
Correction to the last line, it should be,

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_Y X)f$$
 
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  • #6
jbergman
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Correction to the last line, it should be,

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_Y X)f$$
Messed it up again... There must be a better way to remember the formula.
$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_X Y)f$$
 
  • #7
quasar987
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In local coordinates, the hessian of the function ##f## at point ##p## is ##H = \partial_i \partial_k f dx^i \otimes dx^k##. A coordinate-free generalisation is (see) ##H = \nabla df##, or explicitly ##H = \nabla_i (df)_k dx^i \otimes dx^k = \nabla_i \partial_k f dx^i \otimes dx^k##. How is this motivated?
I may be off, and I apologize if I am but my read on this question is that you've overlooked the fact that ##\nabla df## is not just a fancy coordinate-free generalization of the hessian, it is a coordinate independent version of the hessian. I.e. sure you can pick coordinates around a point and look at ##H = \partial_i \partial_k f dx^i \otimes dx^k## but this quantity is meaningless because you can change the coordinates and have it turn into whatever you want. Try it yourself: how does the quantity ##\partial_i \partial_k f dx^i \otimes dx^k## transforms when you look at it from another set of coordinates, say ##(y^i)##?

But the hessian was a pretty good ally... its index at the critical points of f told us about their nature (Morse Lemma). So maybe the hessian is worth fighting for. And by that I mean find a coordinate-independent entity which generalizes the Hessian and maybe share some of its nice properties. We could try looking at ##d^2f## since we know this will be coordinate independent, but no luck, ##d^2=0## always. Turns out we need a connection to make sense of second derivatives. Not too surprising since a connection is the apparatus whose raison d'être is to relate tangent vectors at different points and taking a second derivative involves looking at derivatives defined at different points. So the natural candidate seems to be ##\nabla df##. This a covariant 2-tensor, it is symmetric if ##\nabla## is symmetric and, in local coordinates, it is ##\partial_i \partial_k f dx^i \otimes dx^k## if ##\nabla## is flat.
 
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