The Spin Chronicles Part 32: Ideal exercises

 The Spin Chronicles Part 30:

Solutions to exercises of Part 30

This post is all written by Saša. Saša agreed to present the results of our common work (Saša, Anna, Bjab). So here they are.

Saša with ideals

Solutions to exercises of Part 30

For easier understanding of these exercises and their solutions, let us recall what was said in Part 30 about the self-adjoint idempotents: p = p* = pp = p².

We write a general element of A as p=(p₀,p) (or p=(p,p₄) as in Part 28), where p₀ is a complex scalar, and p is a complex vector. Then p* = (p₀*,p*), where * inside the parenthesis stands for complex conjugation. The p* = p means that p₀* = p₀ and p* = p. In other words p₀ and p must be real.

Now we recall a general multiplication formula for A:
for p=(p₀,p) and q=(q₀,q),
pq = (p₀q₀ + p·q, p₀q + q₀p + i p⨯q).

In particular for pp we get, pp = (p₀² + p·p, 2p₀p) since p⨯p=0.
Thus pp = p implies p₀² + p·p = p₀ and 2p₀p = p.

Then either p=0 or not. If p=0, then p₀² = p₀ and we have two solutions p₀=0 or p₀=1. They correspond to trivial Hermitian idempotents p=0 and p=1. On the other hand, if p is not a zero vector, then from the second equation we get p₀ = 1/2. Substituting this value of p₀ into the first equation we get,
1/4 + p·p = 1/2 or p·p = 1/4.
We deduce that p = 1/2 n, where n is a unit vector in V.

Therefore a general form of a nontrivial Hermitian idempotent is:
p = (1+n)/2 = (1/2,1/2 n).

This nontrivial Hermitian idempotent p = (1/2,1/2 n) will be a crucial element for determining left and right ideals, that these exercises in Part 30 were largely about, as Ark mentioned in his comments to Bjab:

Linear subspace S of A is called left ideal of A if for s in S and any u in A we have that us is in S.
In other words S is invariant under the left action of A. 0 and A are two trivial left ideals.
If there is a non-trivial left ideal, it means that the representation L is reducible (from the definition of reducibility).

and when asked how determining the solution for up=u provides for left ideal,

Let Ip = {s in A: sp=s}.
This is our set. I used the symbol s instead of u. But it is the same set, right?
Take any u in A. Suppose s is in Ip. Is then us also in Ip?
Well, if sp = s, then usp = us, or better (us)p = us.
Therefore us is also in Ip.

So, we are now equipped to start tackling those exercises, and if something additionally turns up to be needed to explain or define, it will be dealt with within the solution of the corresponding exercise.

Exercise 1.
Let n be a unit vector in V. Let T_n denote the space tangent to the unit sphere at n. Thus T_n can be identified with the subspace of V consisting of all vectors of V perpendicular to n.
Let J be defined as a linear operator on T_n, defined by:

Ju = u⨯n.

Show that J is well defined and it defines a complex structure on T_n (i.e. J² = -1).
Show that J is an isometry, that is <Ju,Jv> = <u,v> for all u and v in T_n.

Solution:
From Wikipedia, "In mathematics, a well-defined expression or unambiguous expression is an expression whose definition assigns it a unique interpretation or value."
For n unit vector in V and u a vector in T_n,
Ju = u⨯n = |u| |n| sinφ_(u,n) n_⊥(u,n) = |u| n_⊥(u,n) = u_⊥,
where we used standard definition for cross or vector product "⨯", and that magnitude |n|=1, as well as sinφ_(u,n)=1, because the angle φ_(u,n) between u in T_n and unit vector n is π/2 (90°). The resulting vector u_⊥ has the magnitude |u| and a unique direction given by unit vector n_⊥(u,n) perpendicular to both u and n, thus also in T_n, determined by the right-hand rule for the cross product.
Therefore J is well defined.

J also defines complex structure on T_n, i.e. J² = -1,
J²u = J(Ju) = Ju_⊥ = (u⨯n)⨯n = (n·u)n - (n·n)u = 0u - 1n = -1u,
where in triple product expansion, (a⨯b)⨯c = (c·a)b - (c·b)a, the "·" stands for dot or scalar product which gives 0 for perpendicular vectors.

Regarding the isometry, that is <Ju,Jv> = <u,v>, we recall from Part 28 that our Hilbert space scalar product <u,v> for u=(u₀,u) and v=(v₀,v) in A was defined as scalar part of the product (u₀*,u*)(v₀,v), that is <u,v> = (u₀* v₀ + u*·v), which for u and v in T_n turns into simple dot product (u·v).
<Ju,Jv> = <u⨯n,v⨯n> = <u_⊥,v_⊥> = u_⊥·v_⊥ = |u_⊥| |v_⊥| cosφ_(u⊥,v⊥) = |u| |v| cosφ_(u,v) = u·v = <u,v>,
where angle φ_(u⊥,v⊥) between u_⊥ and v_⊥ is identical to angle φ_(u,v) between u and v, because both u_⊥ and v_⊥ are rotated for the same amount of 90° in the same direction in relation to u and v, respectively.

Exercise 2.
Read Wikipedia article "Linear complex structure", section "Relation to complexifications".
Take J from Exercise 1. and extend it by linearity to T_n^C = T_n + iT_n.
Find a solution of up=u (see the discussion under the previous post Part 29) for p=(1+n)/2 within Tn^C. Express it as an eigenvector of the operator iJ - to which eigenvalue?

Solution:
In the Wikipedia article, we read that,
"If J is a complex structure on V, we may extend J by linearity to V^C:
J(v⊗z) = J(v)⊗z.",
which in case of complex vector u and our J from Exercise 1. means simply,
Ju = u⨯n,
where n is unit vector in V and u complex vector in T_n^C.

Then we proceed with finding a solution of up=u, for p=(1+n)/2=(1/2,1/2 n), where general complex u=(u₀,u) is within T_n^C.
Using the general multiplication formula, for up=u we get:
up = (u₀,u) (1/2,1/2 n) = u = (u₀,u),
(1/2 u₀ + 1/2 u·n, 1/2 u₀n + 1/2 u + i/2 u⨯n) = (u₀,u),
where equating left-hand side with the right-hand one, gives that scalar part u₀ = u·n, while for vector part u, we get the condition:
u = u₀n + i u⨯n.
Since we are looking for a solution to be within T_n^C, that is for those u = u^Re + i u^Im that are perpendicular to n, we see that scalar part u₀ = u·n = 0, which then gives the solution:
u = (u₀,u) = (0,i u⨯n),
that is a complex vector u in T_n^C, u = i u⨯n, which when substituting the expression for J, becomes u = iJu, that is u is eigenvector of iJ with eigenvalue +1.

Cross-check:
up = (0,i u⨯n) (1/2,1/2 n) = u = (0,i u⨯n)
(i/2 (u⨯n)·n, i/2 u⨯n + i i/2 (u⨯n)⨯n) = (0,i u⨯n)
(0, i/2 u⨯n - 1/2 ((n·u)n - (n·n)u) = (0,i u⨯n)
(0, i/2 u⨯n - 1/2 (-u) = (0,i u⨯n)
(0, i/2 u⨯n + i/2 u⨯n) = (0,i u⨯n)
(0,i u⨯n) = (0,i u⨯n).

So, our solution u = iJu = i u⨯n, really is the solution for up=u within T_n^C.

Exercise 3.
Let n and p be as above. Show that up=u if and only if un=u.

Solution:
We are looking that for n unit vector in V and p=(1/2,1/2 n): up=u ⟺ un=u.

Let's first check un=u ⟹ up=u.
For u=(u₀,u) to be solution for un=u, we see that,
un = (u₀,u) (0,n) = u = (u₀,u)
(0 + u·n, 0 + u₀n + i u⨯n) = (u₀,u)
(u·n, u₀n + i u⨯n) = (u₀,u),
scalar part u₀ = u·n and vector part u = u₀n + i u⨯n.
The vector part is therefore composed of the components of which one is proportional to n and the other is perpendicular to n. Then in the scalar part u₀ = u·n evidently contributes only the component that is proportional to n, while for the vector component that is perpendicular to n we saw in Exercise 2. that scalar part is 0, because that general complex vector is in T_n^C. Thus we conclude that the solution of un=u can be decomposed into two independent part as u = u₁ + u₂, where u₁ = z(1,n), z being arbitrary complex scalar or complex number, z = a + ib, and u₂ = i u₂⨯n = u₂, both of which are the solution of un=u on its own.

For u₁=z(1,n):
u₁n = z(1,n)(0,n) = u₁ = z(1,n)
z(0 + n·n,0 + n + i n⨯n) = z(1,n)
z(1,n) = z(1,n)

and for u₂=(0,iu₂⨯n)=(0,u₂):
u₂n = (0,iu₂⨯n)(0,n) = u₂ = (0,u₂)
(0 + i (u₂⨯n)·n,0 + 0 + i (iu₂⨯n)⨯n) = (0,u₂)
(0,-((n·u₂)n - (n·n)u₂) = (0,u₂)
(0,-(0 - u₂)) = (0,u₂)
(0,u₂) = (0,u₂).

Now we need to check if both solutions u₁ and u₂ are also solutions of up=u, where p=(1/2,1/2 n).
For u₂=(0,u₂)=(0,iu₂⨯n) we already saw in Exercise 2. that it is a solution of up=u, so only need to check for u₁=z(1,n):
u₁p = z(1,n)(1/2,1/2 n) = u₁ = z(1,n):
z(1/2 + 1/2 n·n,1/2 n + 1/2 n + 1/2 i n⨯n) = z(1,n)
z(1,n) = z(1,n).

Therefore, un=u ⟹ up=u.

For the other direction, up=u ⟹ un=u, we saw in Exercise 2. that we got the same conditions for the scalar and vector parts of u from up=u like in the first part of this Exercise 3., and that the solution within T_n^C is our u₂=(0,u₂)=(0,iu₂⨯n), and here we also confirmed that u₂ is a solution for un=u. So we would need to check only if u₁ is a solution of u₁p=u₁, which we just did, and that then it satisfies u₁n=u₁, which we also did in the first part of this Exercise.

So, we already proved the other direction, up=u ⟹ un=u, which means that up=u if and only if un=u, or up=u ⟺ un=u.

Exercise 4.
Find an error in the proof below. I can't, but it looks suspicious to me.

Statement.
Let n and n' be two different unit vectors. Then u satisfies both un=u and un'=u if and only if u=0.

Proof.
Set n-n' = v. Suppose v is a nonzero vector. Then we have uv = 0. That means (u·v, u₀v+iu⨯v)=0. Thus u is perpendicular to v. Now, in u₀v+iu⨯v = 0, the first term is proportional to v, the second is perpendicular. Thus both must be zero. It follows that u₀=0, and u⨯v=0. Thus if v is not zero, then u=0.

Solution.
Neither Bjab nor I could spot or find an error in the proof. When asked why the proof looked suspicious, Ark replied in the comments:

This is that two left ideals corresponding to different n's do not intersect (excluding the zero vector). But left and right ideals intersect. It may have some philosophical and physical implications. I will start discussing it in the next post. They are spinor ideals.

A linear subspace S of A is called left ideal of A if for any u in A and s in S we have that us is in S. In other words S is invariant under the left action of A. 0 and A are two trivial left ideals. If there is a non-trivial left ideal that means that the representation L is reducible (from the definition of reducibility). So, we have found an infinite number of ways in which L is reducible - there is a left ideal for each unit vector n in V, and different n's produce different ideals. Spinors, by the standard definition, are elements of such a left ideal. Usually people choose n=e₃.

At this point, it might be useful to show the minimal forms of our two independent solutions of up=u, where p is the general non-trivial Hermitian idempotent p=(1+n)/2. Obviously, the minimal form for u₁=z(1,n) is (1+n), while for u₂=(0,u₂), where u₂=iu₂⨯n is in T_n^C, that is u₂ = u₂^Re + iu₂^Im, we get:
u₂ = u₂^Re + iu₂^Im = i u₂⨯n = i (u₂^Re + iu₂^Im)⨯n =
i (u₂^Re⨯n + iu₂^Im⨯n) = (n⨯u₂^Im + iu₂^Re⨯n),
or u₂^Re = n⨯u₂^Im and u₂^Im = u₂^Re⨯n, which will be handy for next Exercises.

Before going there, we might just notice that really for a given n, we can span the particular left ideal of A by (1+n) and corresponding u₂=iu₂⨯n within T_n^C.

Exercise 5.
Consider the following example:
Choose n = (0,0,1) = e₃. Choose e₁ = (1,0,0) for the real part of u₂. Get the expression for u₂. Define
E₁=(1+n)
E₂ = u₂
Show that E₁ and E₂ span the left ideal of A. For this calculate the action of (e₀,e₁,e₂,e₃) on E₁ and E₂ and express the results as a linear combinations of E₁ and E₂.

Solution.
Choosing n = e₃, we immediately get E₁ = (1+e₃). Then for E₂ = u₂ = u₂^Re + iu₂^Im, after choosing u₂^Re=e₁, u₂^Im = u₂^Re⨯n = e₁⨯e₃ = -e₂, we get E₂ = e₁ - ie₂.

Acting with e₀,e₁,e₂ and e₃ from the left on E₁ and E₂, we get:
e₀(E₁ + E₂) = E₁ + E₂,
e₁(E₁ + E₂) = E₂ + E₁,
e₂(E₁ + E₂) = iE₂ - iE₁,
e₃(E₁ + E₂) = E₁ - E₁,
which can then be written in the matrix form as, for example for e₁:

$\left[\begin{array}{cc}\mathrm{a11}& \mathrm{a12}\\ \mathrm{a21}& \mathrm{a22}\end{array}\right]$ $\left[\begin{array}{c}\mathrm{E1}\\ \mathrm{E2}\end{array}\right]$  =  $\left[\begin{array}{c}\mathrm{E2}\\ \mathrm{E1}\end{array}\right]$ ,

where the resulting matrix [a_ij] transposed, i.e. [a_ij]^T = [a_ji], represents the matrix acting on the basis.

In that way we get:

L(e₀) :  σ₀ = Id = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ ,   L(e₁) :  σ₁ = $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ ,

L(e₂) :  σ₂ = $\left[\begin{array}{cc}0& \mathrm{-i}\\ \mathrm{i}& 0\end{array}\right]$ ,   L(e₃) :  σ₃ = $\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ ,

which are in fact Pauli σ_i matrices.

In an analogous manner we can find the right ideal of A, with the only difference from the procedure for left ideal being that now we look for the solution of pu=u, instead up=u. It is pretty straightforward to get two independent solution of pu=u, one in the non-trivial Hermitian idempotent form of (1+n), and the other with interchanged factors in the cross product compared to the left ideal, u₂=in⨯u₂ being in T_n^C, which then gives u₂^Re = u₂^Im⨯n and u₂^Im = n⨯u₂^Re.

For the calculation of the matrix representation of R(e_i), Ark suggested to choose u₂^Re = e₂, which for the same n = e₃ like in the case of the left ideal of A, u₂^Im = e₃⨯e₂ = -e₁, or for the basis of this particular right ideal of A:
E₁ = (1+n) = (1+e₃)
and
E₂ = u₂ = e₂ - ie₁.

The resulting matrix representation of R(e_i) are again Pauli σ_i matrices, with a slight difference in comparison to the representation L(e_i): σ₂ now corresponded to R(e₁), while σ₁ to R(e₂).

Exercise 6.
Is the representation L of A on our left ideal from Exercise 5. irreducible or reducible?

Solution.
The answer would be "irreducible".
Ark additionally commented:

Representation reducible - the representation space has a non-trivial invariant subspace.
Representation irreducible - no non-trivial invariant subspaces.
Trivial means {0} or the whole space.
Representation space: space on which the representation acts.

In our case: two-dimensional complex space spanned by E₁ and E₂.

That is: our left ideal.

And in Part 31, also said:

Note. In the case we have discussed in previous posts, we have E = A, and, for ρ=L, ρ(u)x=ux. Therefore looking for a non-trivial invariant subspace is the same as looking for a non-trivial left ideal. For ρ=R, the right regular representation, looking for an invariant subspace is the same as looking for a right ideal.

Since we have found non-trivial left and right ideals for A, they suggest that both, left regular representation L(A) and the right one R(A), are in fact reducible, as Anna already reasoned applying the Shur's lemma. Same conclusion can be reached when we look at the bases; in matrix representation, the basis for A was given by 16 4×4 matrices L(e^μ)R(e^ν)=R(e^ν)L(e^μ), that is L(e^μ) and R(e^ν) were represented by 4×4 matrices, while now after finding non-trivial left and right ideals, we see that L(e^μ) and R(e^ν) are represented by 2×2 Pauli σ_i matrices, also suggesting their reducibility.

However, Exercise 6. concerns the reducibility or irreducibility of the representation L of A, that is of the left regular representation of A, on our left ideal, not on the whole A. If it would be reducibile, it would suggest that we could find an invariant subspace where L(e^μ) would be in matrix representation represented by 1×1 matrices, which are in fact just complex scalars, i.e. simple complex numbers. In other words, it would not be a matrix representation anymore.

Work well done

Exercise 1.(simple exercise in abstract thinking) Let A denote the algebra of all 2x2 complex matrices. It is a *-algebra with unit if we define * as the Hermitian conjugate (complex conjugate and transpose), and if we take for 1 the unit matrix. So, we have a particular *-algebra.

Let E denote the space of all complex numbers. Usually we denote it as C. Then E is a one-dimensional complex space. TIn E we have the standard scalar product (x,y). If x,y are two vectors in E, then (x,y)=x*y, the product of complex conjugate of x by y.

Now we define a representation ρ of A on E: ρ(X) = det(X). If X is in A, then det(X) is a complex number. This complex number is interpreted as a linear operator acting on E by multiplication:

ρ(X)x=det(X)x.

Is this a representation?

Is this a *-representation?

Is this representation irreducible?

Saša would probably answer this last question instantly by noticing that since E is -one-dimensional, there are no nontrivial subspaces at all! But I want this last question to be answered by applying the Shur's lemma. 

Exercise 2.(simple exercise in abstract thinking) Let A denote the algebra of all 2x2 complex matrices. It is a *-algebra with unit if we define * as the Hermitian conjugate (complex conjugate and transpose), and if we take for 1 the unit matrix. So, we have a particular *-algebra.

Let E denote the space of all column vectors (a,b), where a ,b are complex numbers. Usually we denote it as C². Then E is a two-dimensional complex space. In E we have the standard scalar product (x,y). Vectors (1,0) and (0,1) form an orthonormal basis in E.

Now we define a representation ρ of A on E: ρ(X) x = Xx.  X is in A, x is in E.

Is this a representation?

Is this a *-representation?

Is this representation irreducible?

I want this last question to be answered by applying the Shur's lemma.