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ergospherical

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- Thread starter ergospherical
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ergospherical

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- #2

martinbn

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How is what motivated?

- #3

ergospherical

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- #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,

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = Y(Xf)-(\nabla_Y X)f$$

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

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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Correction to the last line, it should be,Lastly, it's action on two tangent vectors is given by,

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = Y(Xf)-(\nabla_Y X)f$$

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_Y X)f$$

- #6

jbergman

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Messed it up again... There must be a better way to remember the formula.Correction to the last line, it should be,

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_Y X)f$$

$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_X Y)f$$

- #7

quasar987

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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

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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