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I heard someone talking about how vector fields are the kernels of forms. Can someone give me a detailed explanation about how this works? Thanks.

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kernel? I know vector fields are derivations of forms... –  Ma Ming Apr 30 '13 at 15:19
    
In which sense? The exterior derivative of $n$-forms are $(n+1)$-forms! –  OhMyGod Apr 30 '13 at 15:19
    
When you say kernel, I think of the linear algebra type idea, that an input is in the kernel of a linear map if the corresponding output is zero. Is that what you mean, or something different? –  Muphrid Apr 30 '13 at 15:56
    
@Muphrid it is exactly what I mean. –  OhMyGod Apr 30 '13 at 17:05
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3 Answers

up vote 3 down vote accepted

You might also think about this: vector fields are in the dual space of forms. I'll exemplify this in $\mathbb{R}^3$.

A vector field is a function over a region of three dimensional space that gives at each point a vector. For a point $P=(x,y,z)$ let the the vector field $V$ associates to $P$ the vector $$ V(P) = v_x(P)i_x+v_y(P)i_y+v_z(P)i_z\in\mathbb{R}^3\text{ which is seen here as vector space} $$

The corresponding dual element for $V$ is a function over the region of three dimensional space which gives a linear functional at each point; i.e., $$ w(P) = w_x(P)dx+w_y(P)dy+w_z(P)dz $$ This is called a 1-form, a special case of a differential form. Thus a 1-form is a field of linear functionals.

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If you're in $\mathbb R^n$ (or on a Riemannian manifold) the $1$-form $\omega = \sum F_i dx_i$ naturally corresponds to the vector field $F =(F_1,\dots,F_n)$.

But what I think you have in mind is to associate to $\omega(p)$ the hyperplane it annihilates, i.e., $V_p =\{v\in\mathbb R^n: \omega(p)(v)=0\}=\ker\omega(p)$. When $n=2$, this is a line field on the plane, and so, choosing, for example, a unit vector in each $V_p$ gives us a vector field.

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I'm going to assume that you are working in Euclidean space $\mathbb{R}^n$.

Given a vector space $V$ over $\mathbb{R}$, the dual space $V^*$ is defined to be the space of linear maps $V \to \mathbb{R}$ (check that $V^*$ is in fact a $\mathbb{R}$-vector space).

The vectors at a given point $p \in \mathbb{R}^n$ form a vector field $V_p \cong \mathbb{R}^n$, and a vector field $v$ is just a continuous choice of one vector in each $V_p$. Similarly, a one-form field $\lambda$ is a continuous choice of one vector in each dual space $V_p^*$. So, given a one-form field $\lambda$ and a vector field $v$, we can get an ordinary function as follows: $$ p \mapsto \lambda_p(v_p) \in \mathbb{R} $$

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I think you didn't understand my question... or it wasn't done right. I know what forms and verctor fields are, I want to know how, from one, I can get to the other and vice versa. –  OhMyGod Apr 30 '13 at 15:45
    
A tensor field of type $(1,1)\ldots$ Your question is really badly worded; perhaps try again! –  Rhys Apr 30 '13 at 15:49
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