Diff. forms: M_a = {u /\ a=0 | u in L}

In summary, the conversation discusses exercise 1 of chapter 2 in Flanders' book, which involves proving that the dimension of a subspace is equal to the number of vectors in a wedge product if and only if those vectors are linearly independent. However, the example given in the book is incorrect, as pointed out by the participants in the conversation. The summary also mentions that the expert has comprehended the content and there is no need for further learning.
  • #1
kiuhnm
66
1
Here's exercise 1 of chapter 2 in Flanders' book.
Let ##L## be an ##n##-dimensional space. For each ##p##-vector ##\alpha\neq0## we let ##M_\alpha## be the subspace of ##L## consisting of all vectors ##\sigma## satisfying ##\alpha\wedge\sigma=0##. Prove that ##\dim(M_\alpha)\leq p##. Prove also that ##\dim(M_\alpha)=p## if and only if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##.
(I wrote ##\sigma^1,\ldots## above, but Flanders wrote ##\sigma_1,\ldots## in this exercise. Weird: he uses upper indices in the theory sections.)

My question is this: Isn't it false that if ##\alpha=\sigma^1\wedge\cdots\wedge\sigma^p## where ##\sigma^1,\ldots,\sigma^p## are vectors in ##L##, then ##\dim(M_\alpha)=p##?
 
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  • #2
kiuhnm said:
if α=σ1∧⋯∧σp\alpha=\sigma^1\wedge\cdots\wedge\sigma^p where σ1,…,σp\sigma^1,\ldots,\sigma^p are vectors in LL, then dim(Mα)=p\dim(M_\alpha)=p?
equality ##\alpha\wedge \sigma=0##implies that ##\sigma## belongs to the subspace spanned on ##\{\sigma^1,\ldots,\sigma^p\}##
 
  • #3
wrobel said:
equality ##\alpha\wedge \sigma=0##implies that ##\sigma## belongs to the subspace spanned on ##\{\sigma^1,\ldots,\sigma^p\}##

Let ##\{u^1,u^2,u^3,u^4\}## be a base of ##L##. If we take ##\sigma^1=u^1+u^2## and ##\sigma^2=u^3+u^4## then ##\sigma^1## and ##\sigma^2## are clearly vectors in ##L##, but if ##\alpha=\sigma^1\wedge\sigma^2## then ##\dim(M_\alpha)=1## and not ##2(=p)## as claimed by the text.
The ##\sigma##s also need to be linearly independent!
 
  • #4
The [itex]\sigma^i[/itex] are necessarily independent since [itex]\alpha\neq 0[/itex].

In your example, both [itex]\sigma^1[/itex] and [itex]\sigma^2[/itex] are in [itex]M_\alpha[/itex]. They are independent, so the dimension of [itex]M_\alpha[/itex] is not [itex]1[/itex].
 
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  • #5
kiuhnm said:
Let ##\{u^1,u^2,u^3,u^4\}## be a base of ##L##. If we take ##\sigma^1=u^1+u^2## and ##\sigma^2=u^3+u^4## then ##\sigma^1## and ##\sigma^2## are clearly vectors in ##L##, but if ##\alpha=\sigma^1\wedge\sigma^2## then ##\dim(M_\alpha)=1## and not ##2(=p)## as claimed by the text.

Nope, the dimension is still 2.
 
  • #6
kiuhnm said:
then dim(Mα)=1\dim(M_\alpha)=1 and not 2(=p)2(=p) as claimed by the text.
Thus the textbook is wrong so am i, you have comprehended everything and there is no need for you to learn any more
 
  • #7
wrobel said:
Thus the textbook is wrong so am i, you have comprehanded everything and there is no need for you to learn any more

I thought we were having a civil discussion. Guess I was wrong.
 
  • #8
Infrared said:
The [itex]\sigma^i[/itex] are necessarily independent since [itex]\alpha\neq 0[/itex].

In your example, both [itex]\sigma^1[/itex] and [itex]\sigma^2[/itex] are in [itex]M_\alpha[/itex]. They are independent, so the dimension of [itex]M_\alpha[/itex] is not [itex]1[/itex].

Yes, you're right. Thank you.
 

1. What is the meaning of "Diff. forms" in the expression "M_a = {u /\ a=0 | u in L}"?

"Diff. forms" refers to differential forms, which are mathematical objects used to describe quantities that vary continuously across a space. In this expression, "Diff. forms" refers to a specific type of differential form that satisfies the condition a=0, where "a" is a variable in the form.

2. What does the notation "M_a" represent in the expression "M_a = {u /\ a=0 | u in L}"?

The notation "M_a" represents a set of differential forms that satisfy the condition a=0. This notation is often used in mathematics to represent a set of objects that share a common property or condition.

3. What is the significance of the condition "a=0" in the expression "M_a = {u /\ a=0 | u in L}"?

The condition "a=0" is used to specify a particular type of differential form. This condition ensures that all forms in the set "M_a" have a value of 0 for the variable "a", which can be useful in certain mathematical calculations and applications.

4. How is the set "M_a" related to the set "L" in the expression "M_a = {u /\ a=0 | u in L}"?

The set "M_a" is a subset of the set "L", meaning that all forms in "M_a" are also elements of "L". This implies that the forms in "M_a" satisfy any conditions or properties that are required for elements of "L".

5. Can you provide an example of a differential form that satisfies the condition "a=0" in the expression "M_a = {u /\ a=0 | u in L}"?

One example of a differential form that satisfies the condition "a=0" is the 1-form dx in two dimensions. This form represents the infinitesimal change in the x-coordinate of a point in the plane, and has a value of 0 when a=0. Other examples can include higher dimensional forms such as the 2-form dx ∧ dy in three dimensions, which also has a value of 0 when a=0.

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