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1/21) Wear your math seatbelt, it's time for another differential geometry thread! We'll talk about multivectors and the the cross-product, two concepts linked by the mysterious Hodge star operator!
As usual, I will be shamelessly stealing content from my hero: Roger @penrose !
As usual, I will be shamelessly stealing content from my hero: Roger @penrose !
2/?) We all know of the geometry of (tangent) vectors; they are directional objects. They point somewhere! Physicists like to picture a tangent vector at the point p as the the direction of motion of a possible particle traveling through p.
3/?) Tangent vectors live in the tangent space of all possible directions of motion. This space can also be seen as a very zoomed-out neighborhood of the manifold.
4/?) These tangent spaces are the fundamental building blocks of most of differential geometry! Everything from differential forms to tensors is built from them! And how do you build complex things from simple building blocks? You use algebra!
5/?) Today we will talk about one of the most important of these algebraic operations, the wedge product between two tangent vectors!
The rules are simple:
Antisymmetry:
v ∧ w = - w ∧ v
Linearity:
(a v + b w) ∧ z = a (v ∧ z) + b (w ∧ z)
where a and b are numbers.
The rules are simple:
Antisymmetry:
v ∧ w = - w ∧ v
Linearity:
(a v + b w) ∧ z = a (v ∧ z) + b (w ∧ z)
where a and b are numbers.
6/?) The algebra is simple what what about the geometry? What is this mysterior bi-vector v ∧ w, does it point somewhere? Does it have a geometrical interpretation?
Let's try to use the algebra in the familiar 3D setting!
Let's try to use the algebra in the familiar 3D setting!
7/?) Consider the two 3D vectors
v = x î + y ĵ + z k̂
and
w = a î + b ĵ + c k̂
where î, ĵ and k̂ are cartesian unit vectors. Let's now use our algebraic rules to figure out their wedge products.
v ∧ w = (x î + y ĵ + z k̂) ∧ (a î + b ĵ + c k̂)
v = x î + y ĵ + z k̂
and
w = a î + b ĵ + c k̂
where î, ĵ and k̂ are cartesian unit vectors. Let's now use our algebraic rules to figure out their wedge products.
v ∧ w = (x î + y ĵ + z k̂) ∧ (a î + b ĵ + c k̂)
8/?) First of all, a wedge product of a vector with itself is always zero! This follows from the anti-symmetry:
v ∧ v = - v ∧ v = 0
Remember this as it is the most important property of the wedge product! Now we can use it to simplify our 3D calculation:
v ∧ v = - v ∧ v = 0
Remember this as it is the most important property of the wedge product! Now we can use it to simplify our 3D calculation:
9/?) Algebra time!
v ∧ w = (x î + y ĵ + z k̂) ∧ (a î + b ĵ + c k̂)
= x a (î ∧ î) + x b (î ∧ ĵ) + x c (î ∧ k̂) +
y a (ĵ ∧ î) + y b (ĵ ∧ ĵ) + y c (ĵ ∧ k̂) +
z a (k̂ ∧ î) + z b (k̂ ∧ ĵ) + z c (k̂ ∧ k̂)
v ∧ w = (x î + y ĵ + z k̂) ∧ (a î + b ĵ + c k̂)
= x a (î ∧ î) + x b (î ∧ ĵ) + x c (î ∧ k̂) +
y a (ĵ ∧ î) + y b (ĵ ∧ ĵ) + y c (ĵ ∧ k̂) +
z a (k̂ ∧ î) + z b (k̂ ∧ ĵ) + z c (k̂ ∧ k̂)
10/?) Alright lengthy but easy! Now let's use the anti-symmetry to simplify:
Remember
v ∧ w = - w ∧ v
Therefore:
v ∧ w = x b (î ∧ ĵ) + x c (î ∧ k̂) +
- y a (î ∧ ĵ) + y c (ĵ ∧ k̂) +
- z a (î ∧ k̂) - z b (ĵ ∧ k̂)
Remember
v ∧ w = - w ∧ v
Therefore:
v ∧ w = x b (î ∧ ĵ) + x c (î ∧ k̂) +
- y a (î ∧ ĵ) + y c (ĵ ∧ k̂) +
- z a (î ∧ k̂) - z b (ĵ ∧ k̂)
11/?) Now we just need to collect the terms:
v ∧ w = (x b - y a) (î ∧ ĵ)
+ (x c - z a) (î ∧ k̂)
+ (y c - z b) (ĵ ∧ k̂)
Wait a moment, the coefficients are the components of the cross-product!
v ∧ w = (x b - y a) (î ∧ ĵ)
+ (x c - z a) (î ∧ k̂)
+ (y c - z b) (ĵ ∧ k̂)
Wait a moment, the coefficients are the components of the cross-product!
12/?) This is very helpful as we now can we use our geometric intuition about the cross-product to make sense of the wedge product.
Let's start with a simple mental picture, a bivector v ∧ w can be visualized as the area enclosed by the parallelogram with v and w as sides.
Let's start with a simple mental picture, a bivector v ∧ w can be visualized as the area enclosed by the parallelogram with v and w as sides.
13/?) Using this mental picture the algebra makes more sense. For example, it is clear that v ∧ v is zero since the resulting parallelogram would have zero area. Something is missing however! We need to keep track of the sign, or better of the orientation!
14/?) We need the sign as v ∧ w is clearly different from w ∧ v = - v ∧ w while it has the same associated parallelogram. Does this sign has a geometric interpretation? Yes, you can visualize as a either clockwise or counter-clockwise direction of rotation!
15/?) Note however that without an additional bit of structure the clockwisedness is defined only relatively. You can say whether v ∧ w and z ∧ k have the same or different sign but you can't say that either of which is clockwise.
16/?) The additional piece of structure that you need is nothing less than the strange right hand rule that we all learn in highschool and most of us find arbitrary and strange!
Since it is unlikely that children are reading this I can use one of my favorite xkcd comics!
Since it is unlikely that children are reading this I can use one of my favorite xkcd comics!
17/?) Alright now that we got to the right hand rule we are very close to the cross product! What is missing?
Well the cross-product of two vector is a vector pointing in an orthogonal direction while our bivector is a signed parallelogram! How do we relate these two things?
Well the cross-product of two vector is a vector pointing in an orthogonal direction while our bivector is a signed parallelogram! How do we relate these two things?
18/?) We need two thing. the first thing is a sign, the right hand rule. We say that we need a sign structure on the vector space. The second thing is that we need to define what orthogonal means. If you follow my threads, you'll know that what we need is a metric.
19/?) We can now map our bivector v ∧ w to the unique (metric) vector z that is orthogonal to both v and w, has the magnitude given by the area of the parallelogram and whose sign is given to the right hand rule.
this is implemented by the Hodge star operator:
* (v ∧ w) = z
this is implemented by the Hodge star operator:
* (v ∧ w) = z
20/?) We can now use these facts to write down the result of the star operator on wedge product of unit cartesian bivectors so that the formulas are consistent with the ordinary cross product:
*(î ∧ ĵ) = k̂
*(î ∧ k̂) = ĵ
*(ĵ ∧ k̂) = î
*(î ∧ ĵ) = k̂
*(î ∧ k̂) = ĵ
*(ĵ ∧ k̂) = î
21/21) There would be much more to say but it is now time to stop! If you liked the thread, please like and retweet! Also if you want me to cover a topic you find interesting feel free to ask!
