# tangent vector fields as linear partial differential operators

• May 29th 2011, 03:28 PM
oblixps
tangent vector fields as linear partial differential operators
The vector field X is defined as X(f) = (∂f/∂x_i)v_i where X(x_i) = v_i. Also the directional derivative of f in the direction of v is defined as df(v) = (∂f/∂x_i)v_i. I am using einstein's summation notation so both X(f) and df(v) are sums.

From this we can see that X(f) = df(v). The operator X is defined as
X = (∂/∂x_i)v_i. The source of my confusion comes from the fact that i see df(X) written instead of df(v) where they take ∂/∂x_i to be the basis vectors. I'm confused about why vectors are denoted by ∂/∂x_i which is just an operator? i don't quite understand how an operator X can also act as a vector.

for these operators to be vectors it must be true that they satisfy the properties defining a vector space for all differentiable functions right?

When on a manifold do we always work in this basis of operators?
• May 29th 2011, 11:35 PM
ojones
In differential geometry, tangent vectors are defined as derivations. These are linear maps ${\cal C}^\infty({\cal M})\rightarrow \mathbb{R}$ satisfying Leibniz's rule.

For $\mathbb{R}^n$, the connection between the usual notion of a tangent vector and derivations is provided by the directional derivative.

The reason for viewing tangent vectors as derivations is because the usual notion of a tangent vector doesn't generalize to abstract manifolds.