# Derivations vs Directional derivatives

#### center o bass

In some books, when discussing the relation between partial/directional derivatives and tangent vectors, one makes a generalization called a "derivation". A derivation at $\vec{a} \in \mathbb{R}^n$ is defined as a linear map $D: C^{\infty}(\mathbb{R}^n) \to \mathbb{R}$ which for $f,g \in C^\infty(\mathbb{R}^n)$ that satisfies

$D(fg) = f(a) Dg + g(a) Df.$

On the other hand, some books just stick to directional derivatives, so I wondered: what is the virtue of introducing derivations?

Can someone give an example of something that is a derivation, but not a directional derivative?

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#### Mandelbroth

In some books, when discussing the relation between partial/directional derivatives and tangent vectors, one makes a generalization called a "derivation". A derivation at $\vec{a} \in \mathbb{R}^n$ is defined as a linear map $D: C^{\infty}(\mathbb{R}^n) \to \mathbb{R}$ which for $f,g \in C^\infty(\mathbb{R}^n)$ that satisfies

$D(fg) = f(a) Dg + g(a) Df.$

On the other hand, some books just stick to directional derivatives, so I wondered: what is the virtue of introducing derivations?

Can someone give an example of something that is a derivation, but not a directional derivative?
Derivations are a more general term. One definition of the tangent space at a point $p$ on a manifold $M$ is that the tangent space $T_pM$ is the vector space of all derivations at $p$. However, I can't think of anywhere where you can't think of elements of the tangent space as directional derivatives.

In Lie algebras, the map $\operatorname{ad}_x:\mathfrak{g}\to\mathfrak{g}$ that takes $y\in\mathfrak{g}$ to $[x,y]\in\mathfrak{g}$ (The $[\cdot,\cdot]$ is called the bracket. It's the multiplication operation on $\mathfrak{g}$.) is a derivation, but it isn't a directional derivative.

#### Fredrik

Staff Emeritus
Gold Member
If v is a derivation at p, and $x:U\mapsto\mathbb R^n$ is a coordinate system such that $p\in U$, then there exist real numbers $v^1,\dots,v^n$ such that
$$v=\sum_{k=1}^n v^k\frac{\partial}{\partial x^k}\!\bigg|_p.$$ If this right-hand side is what you consider a directional derivative, then this means that every derivation is a directional derivative.

Not sure if $\mathrm{ad}_x$ should really be considered a derivation, since it's $\mathfrak g$-valued rather than $\mathbb R$-valued.

#### micromass

Not sure if $\mathrm{ad}_x$ should really be considered a derivation, since it's $\mathfrak g$-valued rather than $\mathbb R$-valued.
It's still called a derivation: http://en.wikipedia.org/wiki/Derivation_(differential_algebra [Broken])

Anyway, the Lie bracket can be seen as an actual derivative if you use the notion of Lie derivatives, so its not too farfetched to call it a derivation.

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#### Mandelbroth

It's still called a derivation: http://en.wikipedia.org/wiki/Derivation_(differential_algebra))

Anyway, the Lie bracket can be seen as an actual derivative if you use the notion of Lie derivatives, so its not too farfetched to call it a derivation.
(...ERMAGERD Micro's back!!! )

Micro is correct. There is a construction in differential geometry known as the Lie derivative, which can be applied to various tensor fields. We define the Lie derivative of a vector field $Y$ with $X$ to be $\mathcal{L}_X(Y)=[X,Y]$.

To generalize, for any given noncommutative algebra $A$, the map $D_x: y\mapsto xy-yx$ is a derivation.

Proof: $D_x(yz)=xyz-yzx=xyz-yxz+yxz-yzx=D_x(y)z+yD_x(z)$.