# Why do vectors appear to transform like rank-2 tensors in the Clifford algebra representation?

In Euclidean space with metric $$\delta_{ij}$$ a Galilean vector rotates via $$\hat{V} = \Lambda(\theta) V$$ where $$\Lambda$$ is a member of $$O(3)$$. This can be represented by having the $$V$$ be column vectors and $$\Lambda$$ a $$3\times 3$$ matrix such that $$\Lambda^T=\Lambda^{-1}$$. A second rank symmetric tensor $$T$$ will rotate thus: $$\hat{T} = \Lambda(\theta) T\Lambda^T(\theta)$$. However we can also represent our vectors in a Clifford algebra (as opposed to the representation in the coordinate bases which is typically used).

Given $$\gamma_1,\gamma_2,\gamma_3$$ that satisfy $$\gamma_j\gamma_k+\gamma_k\gamma_j=2\delta_{jk}.$$ And with $$V= v^i\gamma_i$$ The rotation of $$V$$ through an angle $$\theta$$ will be $$\hat{V}= \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) V \exp\left(-\frac{\theta}{2}\gamma_1\gamma_2\right).$$ However this looks an awful lot like the equation for the rotation of a tensor through an angle as well. A similar question arises as to why spinor transformations now look like vector transformations:

$$\hat{\psi}= \exp\left(\frac{\theta}{2}\gamma_1\gamma_2\right) \psi$$ Does employing geometric algebra somehow change our space so that spinors are vectors, and physical vectors, are higher rank tensors in this space?

• You didn't change what things are. Instead you've constructed a representation that acts on spinors instead of vectors. Dec 8, 2019 at 3:48
• Of course, this is just rediscovering tensor products. A rank 2 tensor is the product of two vectors, so its transformation involves two factors of the vector transformation. Similarly, a vector is contained in the product of two spinors. Dec 8, 2019 at 3:54
• Could you elaborate on this? This is surely the thing I'm not understanding Dec 8, 2019 at 3:54
• @Craig $\psi^\dagger\gamma^i\psi$ transforms like a vector Dec 8, 2019 at 3:55
• @knzhou can you suggest more readings on this. All my sources sort of take this for granted. I suppose my question is how this "change of basis" changes our space we're acting on? Dec 8, 2019 at 5:28

In the Clifford algebra framework, the transformation law of spinor $$\psi$$ (a rotation along $$z$$ axis as an example) $$\psi \rightarrow e^{\frac{\theta}{2}\gamma_1\gamma_2} \psi$$ is fundamental, while all other transformations are derived.

Accordingly, a vector $$V^\mu$$ such as fermion current $$V^\mu = \bar{\psi}\gamma^\mu\psi =\psi^\dagger\gamma^0\gamma^\mu\psi,$$ transforms as \begin{align} V^\mu &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(\Lambda^\nu_\mu(\theta)\gamma^\mu) \psi \\ &=\Lambda^\nu_\mu(\theta)\psi^\dagger \gamma^0\gamma^\mu \psi \\ &=\Lambda^\nu_\mu(\theta)V^\mu. \end{align}

Note that

• The fermion current $$V^\mu$$ is defined as $$\bar{\psi}\gamma^\mu\psi= \psi^\dagger\gamma^0\gamma^\mu\psi$$ not as $$\psi^\dagger\gamma^\mu\psi$$. Bear in mind that for Lorentz boost $$(e^{\frac{\theta}{2}\gamma_0\gamma_i})^\dagger\gamma^0 = e^{+\frac{\theta}{2}\gamma_0\gamma_i}\gamma^0 = \gamma^0e^{-\frac{\theta}{2}\gamma_0\gamma_i}$$. The additional $$\gamma^0$$ in $$\bar{\psi} = \psi^\dagger\gamma^0$$ is essential in guaranteeing the Lorentz covariance of the vector $$V^\mu$$.
• $$\Lambda^\nu_\mu(\theta)$$ are just real numbers, so that they can be moved freely around Clifford algebra elements.

And you can also derive that according to the spinor transformation rule, an antisymmetric tensor $$T^{\mu\nu}$$ such as $$T^{\mu\nu} = \bar{\psi}\gamma^\mu\gamma^\nu\psi =\psi^\dagger\gamma^0\gamma^\mu\gamma^\nu\psi,$$ transforms as \begin{align} T^{\mu\nu} &\rightarrow (e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi)^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger(e^{\frac{\theta}{2}\gamma_1\gamma_2})^\dagger\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^0\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2}\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\mu e^{\frac{\theta}{2}\gamma_1\gamma_2} )(e^{-\frac{\theta}{2}\gamma_1\gamma_2}\gamma^\nu e^{\frac{\theta}{2}\gamma_1\gamma_2})\psi \\ &=\psi^\dagger \gamma^0(\Lambda^{\mu'}_\mu(\theta)\gamma^\mu)(\Lambda^{\nu'}_\nu(\theta)\gamma^\nu) \psi \\ &=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)\psi^\dagger \gamma^0\gamma^\mu\gamma^\nu \psi \\ &=\Lambda^{\mu'}_\mu(\theta)\Lambda^{\nu'}_\nu(\theta)T^{\mu\nu} . \end{align}

For related discussions,

• See here for how to embed vectors and tensors in a spinor.
• See here for how to correctly define covariant derivatives with respect to single-sided vs double-sided gauge transformations.