# Spivak's dual space definition?

1. Mar 28, 2008

### Peeter

In calculus on manifolds, within one of the problems of this book the dual space is indirectly defined. I'll quote:

Let $$(R^n)^*$$ denote the dual space of the vector space $$R^n$$. If $$x \in R^n$$, define $$\phi_x \in (R^n)^*$$ by $$\phi_x(y) = \langle x, y\rangle$$.

Now my problem is I just don't get it. Perhaps I have a preconcived idea of dual space that is messing me up (ie: hodge dual, or kernel of a subspace). The dual space description' in this problem doesn't appear to be anything like that.

Note that this is one of the problems in the first chapter of the book and the only sort of inner product discussed has been the $$R^n$$ euclian dot product.

Since only the euclian inner product has been mentioned if one wanted to talk about
$$\phi_x$$ above independent of what it is applied to one can rationalize this as a matrix operator for example:

$$\phi_x(y) = \langle x, y \rangle = x^\text{T} y$$

so $$\phi_x$$ by itself in that sense is just.

$$\phi_x = x^\text{T}$$

(Here $$\phi_x$$ is the matrix of the linear transformation).

If I was to describe the set of all such operators, I'd say that the dual space must then be the set of all the linear operators that map a vector to a number. Is that all this is?

Perhaps somebody can describe for me the application of the dual space as defined by this problem. Some context showing how this is used would help clarify the concept and it's significance.

2. Mar 28, 2008

### Pere Callahan

Indeed.

The dual space of some K-vector space V is the (vector) space of all linear functionals from V to K.

As you pointed out, in the case V=R^n, i.e. whenever V has finite dimension, the dual space V' is equal to V itself - by virtue of the inner product which associates to each element x in R^n a linear functional on R^n (your function phi). Conversely, each linear functional is given in this way (proof?).

For infinite-dimensional vector spaces it is not in general true that the dual space is equal to the space itself.

http://en.wikipedia.org/wiki/Dual_space

Whenever you are confronted with a problem involving linear functionals on some vector space it is good to know the dual (because it is just the space of all linear functionals).

Last edited: Mar 28, 2008
3. Mar 28, 2008

### Peeter

Thanks, that's helpful, and I think I see the connection with hodge dual here now too:

Suppose one defines a subspace as the span of three linearly independent vectors a, b, and c.

Then for a bivector B

$$B = a \wedge b$$

a linear operator of the form:

$$\phi_x(B) = \frac{1}{a \wedge b \wedge c}(x \wedge B)$$

where x is in the span of a,b,c provides a bivector to scalar mapping. Thus, as an operator $$\phi$$ belongs to the dual space for the bivector (and this operator could be applied to any bivector B composed of a wedge product of two vectors in this subspace).

I'm sure that I abusing both notation and language here. It's taken a few months, and many digressions but I'm starting to get enough of an understanding of some of this stuff that I can work my way partially into chapter one of each of my two differential forms books;)

4. Apr 1, 2008

### ObsessiveMathsFreak

This is pretty typical Spivak. Rather than explaining, or in some cases even properly defining something, he lets the reader "discover it for themselves" in an exercise. It means that his books typically take three times longer to read than they should.

Spivak introduces the dual space $$\left(\mathbb{R}^n\right)^*$$ but as mentioned already, neglects to inform us that it is the space of linear functionals. He then introduces the $$\phi_{\mathbf{x}}$$ convention to define some elements of $$\left(\mathbb{R}^n\right)*n$$, but not seemingly all.

However, his next step is to introduce the transformation $$T: \mathbb{R}^n \rightarrow \left(\mathbb{R}^n\right)^* , T(\mathbf{x})=\phi_{\mathbf{x}}$$. He then "invites" the reader to prove that T is a one to one function and conclude that every element in the dual space has the form $$\phi(_{\mathbf{x}})$$, the meaning of which is supposed to be immediately apparent to everyone. I know it wasn't to me.

I would have defined the dual space in the following, more constructive, way.

Let the dual space $$\left(\mathbb{R}^n\right)^*$$ be the space of linear functionals on $$\mathbb{R}^n$$, i.e. all functions $$\phi : \mathbb{R}^n \rightarrow \mathbb{R}$$ such that $$\phi\left( a*\mathbf{x} + b*\mathbf{y}\right)=a\phi\left( \mathbf{x}\right)+b\phi\left(\mathbf{y}\right)$$. So far this space is very abstactly defined. However, we can extract something concrete from this definition.

Let the vectors $$\mathbf{e}_i , i=1\ldotsn$$ be an orthinormal basis for $$\mathbb{R}^n$$. Take $$\mathbf{x}=\Sigma x_i \mathbf{e}_i$$. Then for any element $$\phi \in \left(\mathbb{R}^n\right)^*$$ we have;

$$\phi\left( \mathbf{x}\right)=\phi\left( \Sigma x_i \mathbf{e}_i\right)=\Sigma x_i \phi\left( \mathbf{e}_i\right)$$

So we can say,

$$\phi(\mathbf{x}) = < \mathbf{x}, \Sigma \phi(\mathbf{e}_i)\mathbf{e}_i >$$

Which is effectively how Spivak started, by defining seemingly "some" members of the dual space. This all of course begs the question of how the individual $$\phi(\mathbf{e}_i)$$ are to be calculated in this seemingly circular definition. The answer is that they must be specified initially in some way in order to define $$\phi$$. However, we can be assured that all elements in the dual space can be represented in this form, owing to their linearity.

Hopefully that made some sense. Essentially, the thing to take away here is that every element of the dual space can be thought of as representing a "vector" in an inner product, waiting for another to be applied.

5. Apr 1, 2008

### Peeter

yes, that makes sense.

Your warning about how to approach the book is also good to have (in the intro where he says the problems are the most important part is really not a joke then;)

6. Apr 7, 2008

### wofsy

dual space

A Riemannain metric defines an isomorphism of a vector bundle and its dual bundle of homomorphisms into the base field by sending a vector,v, to the linear map <v,w>.
Each vector w is mapped to a scalar by this inner-product. <v,> is an element of the dual space.

For instance, in Euclidean space, the dual of the differential of a function is its gradient.

The inverse of this map can be used to return the dual vector back. This two way process is used in tensor analysis to "raise and lower" indices.

If one changes the inner product then the dual vector is different. For instance, the usual gradient of a function will not work if the metric is not the usual Euclidean metric. A different gradient will be defined.

To think of a dual vector as merely a vector waiting to inner product with another vector is wrong. Dual vectors exist naturally without an inner product. For instance the differential of a function is defined without an inner product. What the inner product gives is an isomorphism from the vector space and its dual space. This is the key idea, this isomorphism.

Last edited: Apr 7, 2008
7. Apr 7, 2008

### Peeter

I'm not quite sure what your notation means. By gradF here do you mean a directed gradient? What's the w above?

Notation aside, I think I may understand what you are saying and I'd observed the same thing. If one expresses the gradient in terms of a curve parameterized by arc length, the component'' that is multiplied with the unit tangent vector has the form of a chain rule operator. Whereas the parts that are normal (and except in R^2 some of these must be linearly dependent) have a form strikingly similar to the exterior derivative expressed as an operator:

$$\nabla = \underbrace{ \left(\frac{dx_1}{ds}, \frac{dx_2}{ds}, \cdots, \frac{dx_n}{ds}\right) }_{\text{Unit tangent.} } \sum_{i=1}^n \frac{dx_i}{ds}\frac{\partial}{\partial x_i} + \sum_{1 \le i < j \le n} \underbrace{ \left( \mathbf{e}_j \frac{dx_i}{ds} -\mathbf{e}_i \frac{dx_j}{ds} \right) }_{\text{ Normal to unit tangent. }} \left( \frac{dx_i}{ds}\frac{\partial}{\partial x_j} -\frac{dx_j}{ds}\frac{\partial}{\partial x_i} \right)$$

To get this I started with thinking about components of the gradient expressed with reciprocal frame vectors along the curve and a dual, but found this was simpler by just factoring the gradient using the geometric product into the projective and rejective components with respect to the curve.

I thought this was pretty interesting, but don't know yet how this fits in the big picture.

8. Apr 8, 2008

### wofsy

In euclidean space, the gradient,gradF, is just the vector of partial deriviatives (of F), the usual gradient in multivariable calculus. With a non-flat metric you get a different dual and a different gradient.

It has nothing to do with parameterized curves.

9. Apr 8, 2008

### wofsy

If you want to think of the differential in terms of curves, here is the way.

Start with a differentiable function on a manifold and curve on which the manifold is defined. The composition of the function with the curve yields a real valued function of one variable. Take its calculus 1 derivative at a point. This derivative is the value of the differential of the function on the tangent vector to the curve at that point.

This derivative may be calculated in coordinates using the chain rule.

10. Apr 8, 2008

### wofsy

dual spaces

Here is a quick example.

Stand a sphere up on a table and let F be the height of a point on the sphere above the table. F is a nice differentiable function on the sphere.

Let v be any vector in the plane tangent to the north pole. Then dF(v) = 0 because F has a maximum at the north pole. So dF is the dual map which sends each vector to zero. Notice that no inner products were used to calculate dF.

In term of gradients, gradF is normal to the sphere and points in the direction of maximal increase of the height function. dF may also be calculated as the inner product with this normal vector.