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Question on page 7 Flander's book on differential forms

 
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Jan21-10, 03:41 AM   #1
 

Question on page 7 Flander's book on differential forms


On page 7 it gives two conditions for a linear function on the space of p-vectors built from a linear function on the underlying L space. I do not understand! Does anybody ?
Then it continues by saying that the two properties are an axiomatic characterization on the space of p-vectors. So, if I understand correctly, the two properties above for linear functions are true iff the axioms for the p-space on page 5-6 are true? Correct? There's no proof. Does anybody know where I can find it?
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Jan22-10, 09:10 AM   #2
 
Quote by Goldbeetle View Post
On page 7 it gives two conditions for a linear function on the space of p-vectors built from a linear function on the underlying L space. I do not understand! Does anybody ?
Then it continues by saying that the two properties are an axiomatic characterization on the space of p-vectors. So, if I understand correctly, the two properties above for linear functions are true iff the axioms for the p-space on page 5-6 are true? Correct? There's no proof. Does anybody know where I can find it?
The p-vectors form a vector space and satisfy - by definition - certain algebaic properties. These are tersely listed on page 6 - properties (i),(ii), and (iii).

These three properties allow one to characterize linear functions on this vector space in terms of the components of the p-vector. On page 7 he is saying - a linear map on the space of p vectors is a multilinear map on the product space, VxV....V ( p times)
that is also alternating. This you can just prove by calculation.

For instance, take the case of 2 vectors.

Keep in mind that f(as^t + bu^w) = af(s^t) + bf(u^w) because f is linear

f(u^w) = f(-w^u) because u^w = - w^u by (iii)
= -f(w^u) because f is linear. So f viewed as a function,g, of the product space VxV satisfies the relation g(v,w) = -g(w,v) i.e. g is alternating.

f(u^ax + by) = f(u^ax + u^by) = f(au^x + bu^y) by (i) and (iii)
= af(u^x) + bf(u^y) because f is linear. Thus g(u,ax+by) = ag(u,x) + bg(u,y) i.e. g is linear in the second variable. The exact same argument shows that g is also linear in the first variable. Thus g is multilinear.
Jan22-10, 03:17 PM   #3
 
Thanks. What about the axiomatic characterization?
Jan22-10, 03:29 PM   #4
 

Question on page 7 Flander's book on differential forms


Quote by Goldbeetle View Post
Thanks. What about the axiomatic characterization?
Well if you start with a multilinear alternating function then it determines a linear function on p-vectors. So there is a 1-1 correspondence. This means that you can take multilinear and alternating as axioms.
Jan27-10, 04:45 PM   #5
 
Just to comment that this turning of multilinear maps in one space into linear maps
looks like a tensor product of vector spaces.
Jan27-10, 04:55 PM   #6
 
Quote by Bacle View Post
Just to comment that this turning of multilinear maps in one space into linear maps
looks like a tensor product of vector spaces.
The tensor product may be defined exactly in this way.
Jan28-10, 02:42 AM   #7
 
Actually, what Flanders seems to say is that that property of f and g can be used as characterization of p-spaces, that is, for all and only p-spaces f and g are related has he says. Am I right? If so, how come?
Jan28-10, 06:55 AM   #8
 
Quote by Goldbeetle View Post
Actually, what Flanders seems to say is that that property of f and g can be used as characterization of p-spaces, that is, for all and only p-spaces f and g are related has he says. Am I right? If so, how come?
The relation of f and g is called a universal a mapping property.

The p spaces are uniquely determined through the conversion of a multi-linear map on a vector space into a linear map on another vector space.
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