# Coordinate-free Hessian

ergospherical
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?

• lavinia

## Answers and Replies

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?

ergospherical
I'm curious to know why ##H = \nabla df## is the correct generalisation of the usual expression (for local coordinates).

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

Last edited by a moderator:
jbergman
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,

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

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

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