The coordinate-free formulation of the Hessian is expressed as ##H = \nabla df##, which generalizes the local coordinate representation ##H = \partial_i \partial_k f dx^i \otimes dx^k##. This formulation is motivated by the need for a coordinate-independent version of the Hessian, which is crucial for understanding second derivatives in differential geometry. The covariant Hessian, denoted as ##\nabla^2 f##, is defined as ##\nabla^2 f = \nabla(df)## and is symmetric if the connection ##\nabla## is symmetric. The discussion emphasizes the importance of the covariant derivative in establishing a consistent framework for analyzing functions on manifolds.
PREREQUISITESMathematicians, physicists, and students of differential geometry who seek to deepen their understanding of the Hessian and its coordinate-free formulations, particularly in the context of Riemannian manifolds and tensor analysis.
How is what motivated?ergospherical said: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?
Correction to the last line, it should be,jbergman said:Lastly, it's action on two tangent vectors is given by,
$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = Y(Xf)-(\nabla_Y X)f$$
Messed it up again... There must be a better way to remember the formula.jbergman said:Correction to the last line, it should be,
$$\nabla^2f(Y,X) = \nabla^2_{X,Y}f = \nabla_X(\nabla_Y f)-\nabla_{(\nabla_X Y)}f = X(Yf)-(\nabla_Y X)f$$
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)##?ergospherical said: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?
You only need the Leibniz rule and the action of the covariant derivative on scalars and tangent vectors. There is nowhere where metric compatibility enters since cotangent space is the dual of the tangent space. You do not need a metric to do anything of what you describe. Simply definelavinia said:This thread got me thinking about how one might arrive conceptually at the definition of the covariant derivative of a 1 form given an affine connection on the tangent bundle. Here are some thoughts. I apologize in advance for any errors.
One can perhaps motivate the covariant derivative of a 1 form by first taking the case where the affine connection is compatible with a Riemannian metric.
If ##w## is parallel along a curve fitting ##X## and the vector field ##Y## is also parallel then one would want ##w(Y)## to be constant. So the derivative ##X⋅<w_{*},Y>## must equal zero and metric compatibility implies that ##<∇_{X}w_{*},Y>## is also zero. So the dual of ##w## under the Riemannian metric must be parallel along the curve. This suggests that a good definition of the covariant derivative of ##w## is the metric dual of the covariant derivative of its metric dual vector field.
This yields the formula ## (∇_{X}w)(Y)= X⋅w(Y)-w(∇_{X}Y)##
Since this formula does not involve a metric it suggests a definition of covariant derivative of a 1 form for any affine connection.
A similar line of thought might be to start with a set of coordinates for the tangent space at a point and ask how does one extend these coordinates along a curve so that measurement of coordinates of a vector field will not depend on changes in the coordinate 1 forms themselves but only on changes in the vector field. As in standard coordinates in Euclidean space, this can be done by setting the coordinates of the derivative of a vector field to be the derivative of its coordinates. Formally, ##X.w(Y)=w(∇_{X}Y)## for each coordinate 1 form.
When ##w## is not parallel ##X⋅w(Y) - w(∇_{X}Y)## is a 1 form in ##Y## and this again suggests the formula ## (∇_{X}w)(Y)= X⋅w(Y)-w(∇_{X}Y)##
To justify this, it must be shown that ##∇_{X}w## satisfies the definition of a covariant derivative.
Definition
A covariant derivative at a point ##p## on the cotangent bundle of a manifold assigns for each tangent vector ##X_{p}## and each 1 form ##w## a cotangent vector at ##p##. This assignment is linear in both tangent vectors at ##p## and in 1 forms and satisfies the Leibniz rule, ##∇_(X_{p})fw = (X_{p}.f)w_{p} + f∇_{X_{p}}w##. An affine connection is a covariant derivative at each point and is smooth in the sense that the covariant derivatives of a smooth 1 form with respect to a smooth vector field is also a smooth 1 form.
These properties are easily verified.
E.g. by the Leibniz rule, ##(∇_{X}fw)(Y) =X⋅(fw)(Y)- fw(∇_{X}Y)= (X⋅f)w(Y)+f(X⋅w(Y) - fw(∇_{X}Y##
##∇_{X}w## also determines an affine connection on the cotangent bundle.
This is not the case. You are thinking of the case where local coordinates are chosen such that the connection coefficients vanish. While this can be accomplished in a flat space, it is not necessary for all local coordinate systems in a flat space. For example, consider polar coordinates in the two-dimensional Euclidean space.quasar987 said:This a covariant 2-tensor, it is symmetric if ∇ is symmetric and, in local coordinates, it is ∂i∂kfdxi⊗dxk if ∇ is flat.