Hodge Dual as Sequence of Grade Reducing Steps

Click For Summary
SUMMARY

The discussion focuses on establishing a bijection between the exterior powers $$\wedge^p V$$ and $$\wedge^{n-p} V$$ for an inner product space $$V$$. Participants explore the potential of expressing this bijection through a binary function $$G$$, which operates on vectors in a sequence of steps. The dimension of both spaces is confirmed to be $$\binom{n}{p}$$, indicating a vector space isomorphism exists. The conversation also raises the question of whether a map can demonstrate that these spaces maintain the same graded structure.

PREREQUISITES
  • Understanding of exterior algebra and exterior powers, specifically $$\wedge^p V$$ and $$\wedge^{n-p} V$$.
  • Familiarity with vector space isomorphisms and their properties.
  • Knowledge of binary functions and their applications in mathematical mappings.
  • Concept of graded structures in vector spaces.
NEXT STEPS
  • Research the properties of exterior powers in linear algebra.
  • Study the concept of graded structures and their significance in algebraic topology.
  • Explore coordinate-free representations of linear mappings.
  • Investigate applications of binary functions in mathematical operations and transformations.
USEFUL FOR

Mathematicians, graduate students in algebra, and researchers interested in advanced topics in linear algebra and exterior algebra.

MisterX
Messages
758
Reaction score
71
If we seek a bijection $$\wedge^p V \to \wedge^{n-p} V$$ for some inner product space ##V##, we might think of starting with the unit ##n##-vector and removing dimensions associated with the original vector in ##\wedge^p V ##. Might this be expressed as a sequence of steps by some binary function ##G##,
$$\star \left( \mathbf{x} \wedge \mathbf{y} \right) = G\Big(\mathbf{x}, G\big( \mathbf{y}, \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\big)\Big) $$
in which case how might we express ##G##?
 
Physics news on Phys.org
MisterX said:
If we seek a bijection $$\wedge^p V \to \wedge^{n-p} V$$ for some inner product space ##V##, we might think of starting with the unit ##n##-vector and removing dimensions associated with the original vector in ##\wedge^p V ##. Might this be expressed as a sequence of steps by some binary function ##G##,
$$\star \left( \mathbf{x} \wedge \mathbf{y} \right) = G\Big(\mathbf{x}, G\big( \mathbf{y}, \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n\big)\Big) $$
in which case how might we express ##G##?

May I ask why seek a bijection this way? As you probably know, the dimension of ##\wedge^p V## and ##\wedge^{n-p} V## is ##\binom {n} {p}##. Since they already have the same dimension, there is a vector space isomorphism, which you can find in proofs like http://math.stackexchange.com/quest...ctor-spaces-of-equal-dimension-are-isomorphic. A more interesting question, is whether there is a map which shows that ##\wedge^p V## and ##\wedge^{n-p} V## have the same graded structure. I don't know what grade preserving maps are called. Maybe that is what you are trying to find?
 
Lucas SV said:
May I ask why seek a bijection this way? As you probably know, the dimension of ##\wedge^p V## and ##\wedge^{n-p} V## is ##\binom {n} {p}##. Since they already have the same dimension, there is a vector space isomorphism, which you can find in proofs like http://math.stackexchange.com/quest...ctor-spaces-of-equal-dimension-are-isomorphic. A more interesting question, is whether there is a map which shows that ##\wedge^p V## and ##\wedge^{n-p} V## have the same graded structure. I don't know what grade preserving maps are called. Maybe that is what you are trying to find?
Well I was trying to actually express the map (in a coordinate free way), not just prove it exists. But I was also curious if I could find other uses for this operation.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
1K
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K