Proving Decomposability of Forms in Spivak's Book (Vol. 1, Chap. 7)

  • Context: Graduate 
  • Thread starter Thread starter AiRAVATA
  • Start date Start date
  • Tags Tags
    Forms
Click For Summary
SUMMARY

The discussion centers on proving the decomposability of forms in the context of Spivak's "Calculus on Manifolds" (Vol. 1, Chap. 7). It establishes that for a vector space V with dimension less than or equal to 3, every element ω in the second exterior power Λ²(V) is decomposable. The proof utilizes the inner product and the properties of wedge products, demonstrating that any ω can be expressed as a linear combination of wedge products of basis elements from V*. The participants confirm that the approach using the inner product and linearity is valid for establishing the result.

PREREQUISITES
  • Understanding of exterior algebra and the notation of wedge products.
  • Familiarity with vector spaces and their dimensions.
  • Knowledge of linear transformations and inner products.
  • Basic concepts of differential forms as presented in Spivak's work.
NEXT STEPS
  • Study the properties of the exterior algebra in more depth, particularly focusing on decomposability.
  • Learn about the applications of inner products in proving properties of differential forms.
  • Explore the implications of decomposability in higher dimensions and its relation to topology.
  • Review examples of decomposable and non-decomposable forms in various vector spaces.
USEFUL FOR

This discussion is beneficial for mathematics students, particularly those studying differential geometry, algebraic topology, or advanced calculus, as well as educators preparing coursework based on Spivak's texts.

AiRAVATA
Messages
172
Reaction score
0
Sorry to keep bothering, but I am preparing an exam based on Spivak's book on forms (chapter 7 of tome 1).

I need to prove that if \dim V \le 3, then every \omega \in \Lambda^2(V) is decomposable, where an element \omega \in \Lambda^k(V) is decomposable if \omega =\phi_1\wedge\dots\wedge\phi_k for some \phi_i \in V^*=\Lambda^1(V).

I think I must use the inner product, but I am not sure. If \omega \in \Lambda^2(V), then

\omega=a_{12} \phi_1\wedge \phi_2+a_{13}\phi_1\wedge\phi_2+a_{23}\phi_2\wedge \phi_3

I know that if \{v_1,v_2,v_3\} are a basis of V, then
\begin{array}{l} i_{v_1}\phi_1\wedge\phi_2\wedge\phi_3=\phi_2\wedge\phi_3 \\<br /> i_{v_2}\phi_1\wedge\phi_2\wedge\phi_3=-\phi_1\wedge\phi_3 \\<br /> i_{v_3}\phi_1\wedge\phi_2\wedge\phi_3=\phi_1\wedge\phi_2<br /> \end{array}

so

\omega=(a_{12}i_{v_3}-a_{13}i_{v_2}+a_{23}i_{v_1}) \phi_1\wedge\phi_2\wedge\phi_3

and given the linearity

\omega=i_v\phi_1\wedge\phi_2\wedge\phi_3

where v=a_{21}v_1-a_{13}v_2+a_{12}v_3.

Does that prove the result?

Other idea I had is to express \phi_i in terms of the base of \Lambda^1(V), but I seem to going nowhere.
 
Physics news on Phys.org
I think that I've got it.

Let \phi_1,\phi_2\in \Lambda^1(V), where
\begin{array}{l}<br /> \phi_1=a_1\varphi_1+a_2\varphi_2+a_3\varphi_3 \\<br /> \phi_2=b_1\varphi_1+b_2\varphi_2+b_3\varphi_3<br /> \end{array}<br />

Then

\phi_1\wedge \phi_2=(a_1b_2-a_2b_1)\varphi_1\wedge\varphi_2+ (a_1b_3-a_3b_1)\varphi_1\wedge\varphi_3+ (a_2b_3-a_3b_2)\varphi_2\wedge\varphi_3[/tex]<br /> <br /> So, given \omega \in \Lambda^2(V), there are (many?) \phi_1,\phi_2\in\Lambda^1(V) such that \omega=\phi_1\wedge\phi_2.<br /> <br /> What do you guys think?
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K