Proof set of one-forms is a vector space

[Azrael84]

Hi,

I am currently working through 'Schutz-First course in General Relativity' problem sets. Question 2 of Chapter 3, asks me to prove the set of one forms is a vector space.

Earlier in the chapter, he defines:

$$\tilde{s}=\tilde{p}+\tilde{q}$$
$$\tilde{r}=\alpha \tilde{p}$$

To be the one forms whose values on a vector $$\vec{A}$$ are:


$$\tilde{s} (\vec{A})=\tilde{p}(\vec{A})+\tilde{q}(\vec{A})$$
$$\tilde{r}(\vec{A})=\alpha \tilde{p}(\vec{A})$$

Given this definition, surely some of the axioms to be a vector space, are simply satisfied by definition (namely closure under addittion, and closure under scalar multiplication), so how does one prove these things, if they have been definied so? Or should I be looking to prove other vector space axioms, like existence of additive inverse etc? If so still not sure where to start.

Any help, much appreciated.

Thanks

 

[matheinste]

Hi,

I am currently working through 'Schutz-First course in General Relativity' problem sets. Question 2 of Chapter 3, asks me to prove the set of one forms is a vector space.

Earlier in the chapter, he defines:

$$\tilde{s}=\tilde{p}+\tilde{q}$$
$$\tilde{r}=\alpha \tilde{p}$$

To be the one forms whose values on a vector $$\vec{A}$$ are:


$$\tilde{s} (\vec{A})=\tilde{p}(\vec{A})+\tilde{q}(\vec{A})$$
$$\tilde{r}(\vec{A})=\alpha \tilde{p}(\vec{A})$$

Given this definition, surely some of the axioms to be a vector space, are simply satisfied by definition (namely closure under addittion, and closure under scalar multiplication), so how does one prove these things, if they have been definied so? Or should I be looking to prove other vector space axioms, like existence of additive inverse etc? If so still not sure where to start.

Any help, much appreciated.

Thanks

Just start with the definitions/axioms given for a vector space and put one forms in place of vectors. If one forms satisfy all the definitions/axioms then they form a vector space. As you pointed out the basic behaviour of one forms is defined to be the same as vectors and so they satisfy the vector space axioms by definition.

Matheinste.

 

[Azrael84]

Just start with the definitions/axioms given for a vector space and put one forms in place of vectors. If one forms satisfy all the definitions/axioms then they form a vector space. As you pointed out the basic behaviour of one forms is defined to be the same as vectors and so they satisfy the vector space axioms by definition.

Matheinste.

I am aware that to prove something is a vector space, you simply show that it satisfies the axioms of a vector space.

The problem is like you say, the basic behaviour of one-forms has been defined to satisfy these axioms by Schutz earlier in the chapter. So I do not understand what he is lucking for when he asks me to prove one-forms form a vector space.

For example how would you prove that one-forms satisy closure under addition?

 

[matheinste]

I am aware that to prove something is a vector space, you simply show that it satisfies the axioms of a vector space.

The problem is like you say, the basic behaviour of one-forms has been defined to satisfy these axioms by Schutz earlier in the chapter. So I do not understand what he is lucking for when he asks me to prove one-forms form a vector space.

For example how would you prove that one-forms satisy closure under addition?

Check, using the vector space axiom for addition, that adding two arbitrary one forms produces a one form. They do so by definition of one form addition as given in the text. I'm afraid this may not answer your question but I don't know how else to put it.

Matheinste

 

[Azrael84]

I can't help but feel he is after something more.

$$\tilde{p}(\vec{A})+\tilde{q}(\vec{A})$$
$$= \tilde{p}_0\vec{A}^0+\tilde{q}_0\vec{A}^0+ \tilde{p}_1\vec{A}^1+\tilde{q}_1\vec{A}^1+...<br /> =(\tilde{p}_0+\tilde{q}_0)\vec{A}^0+ (\tilde{p}_1+\tilde{q}_1)\vec{A}^1+...$$

Which if we define s=p+q, to mean:
$$\tilde{s}_0=\tilde{p}_0+\tilde{q}_0, <br /> \tilde{s}_1=\tilde{p}_1+\tilde{q}_1$$

Would show that p+q is still a one form.

All seems rather circular and hollow anyhow, not sure if there is a better way.

 

[hamster143]

You are essentially correct. Like many theorems in basic math, it is much easier to prove it than to formulate accurately what we know and what we need to prove.

Definition 1. A one-form is a linear function mapping vectors into real numbers. Linear means

$$\tilde{p}(a\vec{A}+b\vec{B}) = a\tilde{p}(\vec{A})+bp(\vec{B})$$

Definition 2. We define operations of addition and multiplication by a scalar for a one-form.$$\tilde{s}=\tilde{p}+\tilde{q}$$
$$\tilde{r}=\alpha \tilde{p}$$

To be the one forms whose values on a vector $$\vec{A}$$ are:

$$\tilde{s} (\vec{A})=\tilde{p}(\vec{A})+\tilde{q}(\vec{A})$$
$$\tilde{r}(\vec{A})=\alpha \tilde{p}(\vec{A})$$

That's ALL we know about one-forms at this point. We're asked to prove on the basis of this knowledge that the space of one-forms is a vector space (defined on page 342). Namely, that it is an abelian group and that multiplication of a one-form by a scalar is sufficiently well-behaved. To do that, we need to "prove" a large number of trivial properties such as that addition of one-forms is commutative (a+b=b+a), most of which follow immediately from the definition 2.

One part of the "proof" that requires more than checking the definition is to prove that the space of one-forms is, in fact, closed under addition and multiplication by scalar. To prove that, we need to prove that the object we make by adding two arbitrary one-forms or multiplying a one-form by a number according to the definition is still a one-form (it is still linear in its argument). But even that is very easy to show.

 

[Azrael84]

Thanks Hamster. I think Def 1 is definitely key to this like you say, with this definition of what a one form must do to be a one-form, you can test that the new object (via say addittion of two one forms) is still a one form, etc.

thanks