HW 10 Paul Lea

Problem 1 N9.3 The contribution to the anomaly by a given irreducible representation is determined by the trace of the product of a generator and the anti-commutator of two generators, namely, $\mathrm{Tr}{T}^A\{T^B{T}^C\}$ Here T A denotes a generator of the gauge group G. Show that for G = SU (5), the anomaly cancels between the $5^{\ast }$ and the 10. Note that for SU (N), we can, with no loss of generality, take A, B, and C to be the same, so that the anomaly is determined by the trace of a generator cubed (namely, $\mathrm{Tr}{T}^3$ ). It is rare that we get something cubed in physics, and so any cancellation between irreducible representations can hardly be accidental.

Easiest to use diagonal generator.

$$T=\text{diag}\left(\frac{1}{3},\frac{1}{3},\frac{1}{3},-\frac{1}{2},-\frac{1}{2}\right)$$

Taking this diagonal generator, we evaluate the trace of the generator cubed in the 5* dimensional representation.

All entries are real: (multiply by 6)

$$T^{\ast }=\text{diag}\left(2,2,2,-3,-3\right)$$

Cubing this matrix:

$$T^{\ast 3}=\text{diag}\left(8,8,8,-27,-27\right)$$

Taking the trace:

$$\mathrm{Tr}(T^{\ast 3})=-30$$

Now for the 10-dimensional representation: Basis for 10-d representation:

$$(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)$$

Converting the 5-d diagonal matrix to a 10-d representation

$$T_{\text{10 d}}=\text{diag}(d_1+d_2,d_1+d_3,d_1+d_4,d_1+d_5,d_2+d_3,d_2+d_4,d_2+d_5,d_3+d_4,d_3+d_5,d_4+d_5)$$ $$T_{\text{10 d}}=\text{diag}(4,4,-1,-1,4,-1,-1,-1,-1,-6)$$

Cubing this diagonal matrix:

$$T_{\text{10 d}}^3=\text{diag}(64,64,-1,-1,64,-1,-1,-1,-1,-216)$$

Taking the trace:

$$\mathrm{Tr}{T}_{\text{10 d}}^3=3(64)-222=192-222=-30$$ $$30-(-30)=0$$

Problem 3 N9.3 Work out how the 3-indexed antisymmetric 10 dimensional tensor in SO(10) decomposes on restriction to $SO(4)\otimes SO(6)$.

• 3 indexed antisymmetric 10-d tensor has 120 elements.

• Symmetric tensor rep of SO(6) has 20 elements $\frac{1}{2}(6)(7)-1$

• Adjoint rep of SO(6) has 15 elements $\frac{1}{2}(6)(5)$

• The vector representation of SO(6) has 6 (obviously)

• Trivial representation has 1

• The vector representation of SO(4) has 4 elements

• The trivial representation has 1 element

• Tensor product representations (3,1) and (1,3)

Decomposing Using the following equation (credit Nate Laposky and Hannah Turner)

$${\Lambda }^3(V\oplus W)={\Lambda }^3(V)\oplus \left({\Lambda }^2(V)\otimes W\right)\oplus \left(V\otimes {\Lambda }^2(W)\right)\oplus {\Lambda }^3(W)$$

We can plug in our representations:

$${\Lambda }^3(10)={\Lambda }^3(4,1)\oplus \left({\Lambda }^2(4,1)\otimes (1,6)\right)\oplus \left((4,1)\otimes {\Lambda }^2(1,6)\right)\oplus {\Lambda }^3(1,6)$$

And from this we can start pulling out our representations: We have a (4,1) representation from ${\Lambda }^3(4,1)$

From $\left({\Lambda }^2(4,1)\otimes (1,6)\right)$ we get

$$(6,6)$$

From $\left((4,1)\otimes {\Lambda }^2(1,6)\right)$ we get

$${\Lambda }^2(6)\cong 15\to (4,15)$$

From ${\Lambda }^3(1,6)$ we get the 20 dimensional representation of $SO(6)$

$$(1,20)$$

Decomposing Using the following equation (credit Nate Laposky and Hannah Turner)

$${\Lambda }^3(V\oplus W)={\Lambda }^3(V)\oplus \left({\Lambda }^2(V)\otimes W\right)\oplus \left(V\otimes {\Lambda }^2(W)\right)\oplus {\Lambda }^3(W)$$

We can plug in our representations:

$${\Lambda }^3(10)={\Lambda }^3(4,1)\oplus \left({\Lambda }^2(4,1)\otimes (1,6)\right)\oplus \left((4,1)\otimes {\Lambda }^2(1,6)\right)\oplus {\Lambda }^3(1,6)$$

And from this we can start pulling out our representations: We have a (4,1) representation from ${\Lambda }^3(4,1)$

From $\left({\Lambda }^2(4,1)\otimes (1,6)\right)$ we get

$$(6,6)$$

From $\left((4,1)\otimes {\Lambda }^2(1,6)\right)$ we get

$${\Lambda }^2(6)\cong 15\to (4,15)$$

From ${\Lambda }^3(1,6)$ we get the 20 dimensional representation of $SO(6)$

$$(1,20)$$

Taking this collection of representations, we can

$$120\to (20,1)\oplus (15,4)\oplus (6,6)\oplus (1,4)$$ $$\to (20,1)\oplus (15,4)\oplus (6,3)\oplus (6,3)\oplus (1,4)$$