Spinor Decomposition of \((16^+)\otimes(16^+)\) in SO(10)

Five explanations on \((16^+)\otimes(16^+)\), antisymmetric tensors, self-duality, and differential forms.

1. Decomposition of \((16^+)\otimes(16^+)\)

The idea is that the tensor product of two chiral spinors can be expanded in a basis built from antisymmetrized products of gamma matrices, and the chirality projectors tell you which ranks survive.

For \(SO(10)\), we have \(2n=10\), so \(n=5\). The Dirac spinor has dimension \(2^5=32\), and it splits into two Weyl spinors

32 = 16^+ \oplus 16^-.

We want to understand

16^+ \otimes 16^+.

1.1 Start from the gamma-matrix expansion

Given two spinors, a general bilinear can be written using

C\Gamma_{[k]},

where \(\Gamma_{[k]}\) means the antisymmetrized product of \(k\) gamma matrices:

\Gamma_{[k]}=\Gamma_{i_1i_2\cdots i_k} \equiv \Gamma_{[i_1}\Gamma_{i_2}\cdots \Gamma_{i_k]}.

So the tensor product of two spinors decomposes into antisymmetric tensor representations \([k]\), where \([k]\) denotes the rank-\(k\) antisymmetric tensor of \(SO(10)\).

In \(10\) dimensions, these have dimensions

[0]:1,\quad [1]:10,\quad [2]:45,\quad [3]:120,\quad [4]:210,\quad [5]:252.

1.2 Use chirality

Let

P_\pm=\frac12(1\pm \Gamma_F),

with \(\Gamma_F\) the chirality matrix. For a positive-chirality spinor,

\psi_R=P_+\psi.

Now consider the bilinear

\psi^T P_+\, C\, \Gamma_{[k]}\, P_+ \psi.

For \(n\) odd one has

\[ P_+ C = C P_-. \]

Also, \(\Gamma_F\) anticommutes with each gamma matrix, so with a product of \(k\) gamma matrices it gives

\Gamma_F \Gamma_{[k]} = (-1)^k \Gamma_{[k]} \Gamma_F.

That means:

if \(k\) is odd, \(\Gamma_{[k]}\) flips chirality, so \(\Gamma_{[k]} P_+ = P_- \Gamma_{[k]}\);
if \(k\) is even, \(\Gamma_{[k]}\) preserves chirality, so \(\Gamma_{[k]} P_+ = P_+ \Gamma_{[k]}\).

Therefore:

\psi^T P_+ C\Gamma_{[k]}P_+\psi = \psi^T C P_- \Gamma_{[k]} P_+\psi.

Now move \(P_-\) through \(\Gamma_{[k]}\):

for \(k\) odd, \(P_- \Gamma_{[k]} = \Gamma_{[k]} P_+\), so the expression is nonzero in general;
for \(k\) even, \(P_- \Gamma_{[k]} = \Gamma_{[k]} P_-\), so one gets \(P_-P_+=0\).

So for \(n=5\) odd, only odd \(k\) survive in

16^+\otimes16^+.

Thus at first sight,

16^+\otimes16^+ = [1]\oplus [3]\oplus [5].

1.3 Dimensions of the surviving pieces

For \(SO(10)\),

\dim [1] = \binom{10}{1}=10,

\dim [3] = \binom{10}{3}=120,

\dim [5] = \binom{10}{5}=252.

But

16^+\otimes16^+

has dimension

16\times 16=256.

If we added the full \([5]\), we would get

\[ 10+120+252=382, \]

which is too large. So something still has to happen to \([5]\).

1.4 The rank-5 tensor splits into self-dual and anti-self-dual parts

In \(10=2n\) dimensions, an \(n\)-form can be Hodge dualized into another \(n\)-form. So a 5-form in ten dimensions splits into two irreducible pieces:

[5] = [5]_+ \oplus [5]_-,

each of dimension

\frac{252}{2}=126.

These are the self-dual and anti-self-dual 5-forms.

For two spinors of the same chirality, only one of these two halves appears. Hence the actual decomposition is

16^+\otimes16^+ = [1]\oplus[3]\oplus[5]_+,

or, in dimensions,

16^+\otimes16^+ = 10\oplus 120\oplus 126.

1.5 Dimension check

10+120+126=256=16\times16.

1.6 Why the 5-form is only half of the full 252

For \(k=n=5\), the gamma-matrix identity involving the chirality matrix relates \(\Gamma_{[5]}\) to its Hodge dual. Schematically,

\Gamma_{i_1\cdots i_5} \sim \epsilon_{i_1\cdots i_5 j_1\cdots j_5}\, \Gamma^{j_1\cdots j_5}\Gamma_F.

When this acts on a spinor of definite chirality, \(\Gamma_F=\pm 1\), so the 5-form piece automatically satisfies a self-duality or anti-self-duality condition. Thus only one \(126\) survives.

Final decomposition: \[ 16^+ \otimes 16^+ = 10\oplus120\oplus126. \]

2. How to Calculate the Number \(120\)

The \(120\) comes from the dimension of the rank-3 antisymmetric tensor representation of \(SO(10)\).

A rank-3 antisymmetric tensor has components

T_{[abc]}, \qquad a,b,c=1,\dots,10,

with complete antisymmetry:

T_{abc}=-T_{bac}=-T_{acb}, \quad \text{etc.}

Because of antisymmetry, the only independent components are those with three distinct indices, and each unordered choice of 3 indices gives exactly one independent component.

So the number of independent components is just the number of ways of choosing 3 distinct indices from 10:

\dim[3]=\binom{10}{3}.

Now compute it:

\binom{10}{3} = \frac{10!}{3!\,7!} = \frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1} = \frac{720}{6} = 120.

\boxed{\dim[3]=120.}

That is why the \([3]\) piece in

16^+\otimes16^+= [1]\oplus[3]\oplus[5]_+

has dimension \(120\).

For comparison:

\dim[1]=\binom{10}{1}=10,

\dim[5]=\binom{10}{5}=252,

and then the self-dual half gives

\[ 252/2=126. \]

So the full count is

10+120+126=256=16\times16.

3. Why the Rank-5 Tensor Splits into Self-Dual and Anti-Self-Dual Parts

This is one of the key geometric facts behind the \(126\) in \(SO(10)\).

3.1 What is a rank-5 antisymmetric tensor?

In \(10\) dimensions, a rank-5 antisymmetric tensor is

T_{i_1 i_2 i_3 i_4 i_5},

totally antisymmetric in its indices.

The number of independent components is

\binom{10}{5}=252.

3.2 The Hodge dual

In even dimensions, there is a natural operation called the Hodge dual.

It maps a \(k\)-form into a \((10-k)\)-form using the Levi-Civita tensor:

(*T)_{j_1\cdots j_{10-k}} = \frac{1}{k!}\, \epsilon_{j_1\cdots j_{10-k} i_1\cdots i_k} \,T^{i_1\cdots i_k}.

3.3 Special feature when \(k=5\)

In \(10\) dimensions, if \(k=5\), then \(10-k=5\). So the Hodge dual maps

\boxed{5\text{-form} \;\longrightarrow\; 5\text{-form}}

That means the duality operation acts within the same space.

3.4 Applying the dual twice

If you apply the Hodge dual twice, you get (in Euclidean signature)

\[ *(*T) = T. \]

So the dual operator satisfies

\[ *^2 = 1. \]

This is exactly like an operator with eigenvalues

\boxed{+1 \quad \text{or} \quad -1.}

3.5 Eigenvectors of the dual: self-dual vs anti-self-dual

Since \(*^2 = 1\), we can decompose any 5-form into eigenvectors of \(*\):

Self-dual part

\[ *T = +T \]

Anti-self-dual part

\[ *T = -T \]

Every tensor splits uniquely as

\[ T = T_+ + T_-, \]

where

T_\pm = \frac{1}{2}(T \pm *T).

3.6 Why the dimension halves

The Hodge dual is a linear operator acting on a 252-dimensional space, with eigenvalues \(+1\) and \(-1\).

Because it has equal numbers of \(+1\) and \(-1\) eigenvalues, the space splits evenly:

\[ 252 = 126 + 126. \]

So:

\boxed{[5] = [5]_+ \oplus [5]_- = 126 \oplus 126}

3.7 Intuition

A 5-form in 10D is “halfway” to being dual to itself. Duality pairs each component with another one, but at exactly half dimension the pairing becomes a constraint, cutting the degrees of freedom in half.

3.8 Why only one \(126\) appears in \(16^+ \otimes 16^+\)

There is a key identity:

\Gamma_{[5]} \sim \epsilon \,\Gamma_{[5]} \Gamma_F.

When acting on a chiral spinor,

\Gamma_F = \pm 1,

so this becomes a self-duality condition on the tensor constructed from the spinors. Thus:

\(16^+ \otimes 16^+\) picks one of the two halves,
not both.

3.9 Analogy

This is very similar to electromagnetism in 4D:

the field strength \(F_{\mu\nu}\) has 6 components;
it splits into self-dual (3 components) and anti-self-dual (3 components).

4. What Is a \(k\)-Form?

A \(k\)-form is a completely antisymmetric tensor with \(k\) indices. It is best understood as something you can integrate over a \(k\)-dimensional surface.

4.1 Basic definition

A \(k\)-form \(\omega\) in \(n\) dimensions has components

\omega_{i_1 i_2 \cdots i_k}

that are totally antisymmetric:

\omega_{i_1 i_2 \cdots i_k} = -\omega_{i_2 i_1 \cdots i_k} = \cdots

So swapping any two indices flips the sign.

4.2 Simple examples

0-form: just a scalar function

\phi(x)

1-form: like a covector

\[ A_i \]

written as

A = A_i\, dx^i

2-form:

B_{ij} = -B_{ji}

written as

B = \tfrac{1}{2} B_{ij}\, dx^i \wedge dx^j

General \(k\)-form:

\omega = \frac{1}{k!}\,\omega_{i_1 \cdots i_k}\, dx^{i_1} \wedge \cdots \wedge dx^{i_k}

4.3 The wedge product

The symbol \(\wedge\) means antisymmetric product. For example,

dx^i \wedge dx^j = - dx^j \wedge dx^i

so automatically

dx^i \wedge dx^i = 0.

4.4 Counting components

Because of antisymmetry, the number of independent components is

\boxed{\binom{n}{k}}.

Example in 10D:

1-form: \(10\)
2-form: \(45\)
3-form: \(120\)
5-form: \(252\)

4.5 Why they matter in physics

\(k\)-forms naturally describe physical objects:

1-forms \(\to\) gauge fields \(A_\mu\)
2-forms \(\to\) electromagnetic field \(F_{\mu\nu}\)
higher forms \(\to\) string theory and supergravity objects

4.6 Geometric meaning

A \(k\)-form is something you integrate over a \(k\)-dimensional object:

Form	Integrates over
0-form	points
1-form	curves
2-form	surfaces
3-form	volumes
\(k\)-form	\(k\)-dimensional surfaces

4.7 Connection to the \(SO(10)\) problem

When we write

[3],\; [5]

these are exactly the 3-form and 5-form representations. So

\Gamma_{[k]} \leftrightarrow \text{\(k\)-form structure}.

That is why the decomposition

16^+ \otimes 16^+ = [1]\oplus[3]\oplus[5]_+

is really saying: vector (1-form), 3-form, self-dual 5-form.

One-line summary: \[ \boxed{\text{A \(k\)-form is a totally antisymmetric rank-\(k\) tensor that lives naturally on \(k\)-dimensional geometry.}} \]

5. Illustrating the Counting of Components for the 2-Form and 3-Form

Let us explicitly count the independent components so you can see where the binomial coefficients come from.

5.1 2-form in \(n\) dimensions

A 2-form is

B_{ij} = -B_{ji}.

Step 1: Start with all components. There are \(n^2\) components \(B_{ij}\).

Step 2: Impose antisymmetry.

Diagonal terms vanish: \(B_{ii}=0\).
Off-diagonal terms are related: \(B_{ij}=-B_{ji}\).

So each pair \((i,j)\) with \(i\neq j\) gives only one independent component.

Step 3: Count independent pairs. We only count combinations with

\[ i < j. \]

Number of such pairs:

\binom{n}{2} = \frac{n(n-1)}{2}.

Example: \(n=4\)

Independent components:

B_{12},\; B_{13},\; B_{14},\; B_{23},\; B_{24},\; B_{34}

Total:

\boxed{6 = \binom{4}{2}}.

5.2 3-form in \(n\) dimensions

A 3-form is:

C_{ijk} = \text{totally antisymmetric}.

Step 1: Antisymmetry rules.

Swapping any two indices flips sign.
Any repeated index gives zero: \(C_{iik}=0\).

So only triples with all indices distinct survive.

Step 2: Count independent triples. We only count ordered sets

\[ i < j < k. \]

Each such triple corresponds to exactly one independent component.

Number of choices:

\binom{n}{3} = \frac{n(n-1)(n-2)}{6}.

Example: \(n=4\)

Independent components:

C_{123},\; C_{124},\; C_{134},\; C_{234}

Total:

\boxed{4 = \binom{4}{3}}.

5.3 Why this works

Antisymmetry means:

order does not matter, up to a sign;
repetitions vanish.

So each independent component corresponds exactly to choosing a set of \(k\) distinct indices, that is, a combination.

5.4 Visual intuition

Think of it like this:

2-form \(\to\) choose 2 directions \(\to\) oriented area element
3-form \(\to\) choose 3 directions \(\to\) oriented volume element

Each distinct choice of directions gives one independent component.

5.5 Connection to the \(SO(10)\) case

Now you can immediately see:

\binom{10}{2} = 45, \qquad \binom{10}{3} = 120,

which is exactly where the \(120\) in

16^+ \otimes 16^+ = 10 \oplus 120 \oplus 126

comes from.

Summary: a 2-form counts pairs i < j, a 3-form counts triples i < j < k, and antisymmetry removes redundancy while killing repeated indices.

This is additional material to complement page 417 of Group Theory in a Nutshell by Anthony Zee.

Created with assistance from ChatGPT.