Is spinor $\psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?

Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $\omega$ in Einstein-Cartan gravity $$ \omega = \frac{1}{4}\omega^{ab}_{\mu}\gamma_a\gamma_bdx^{\mu}. $$ The spin connection 1-form $\omega$ is a

  1. Vector in spacetime manifold characterized by differential vector base $dx^{\mu}$.
  2. Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $\gamma_a\gamma_b$, where $\gamma_a$ are Dirac vector basis.

(A bonus example: the torsion 2-form $T = \frac{1}{2}T^{a}_{\mu\nu}\gamma_a dx^{\mu}dx^{\nu}$ is a bi-vector and vector in the above two spaces, respectively)

Now back to our initial assertion: a spinor is a

  1. Scalar in spacetime manifold characterized by differential vector base $dx^{\mu}$. The spinor $\psi$ is a $0$-form.
  2. Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $\gamma_a$: $\psi = \xi_s + \xi_v^{a}\gamma_a + \xi_{bv}^{ab}\gamma_a\gamma_b + ...$. The individual coefficients like $\xi_s$ and $\xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as an isospin doublet) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(\bar{\psi}\psi)$, vector bi-linear (current) $tr(\bar{\psi}\gamma_a\psi)$, bi-vector bi-linear $tr(\bar{\psi}\gamma_a\gamma_b\psi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac algebra. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor. There are only 2 column spinors (electron/neutrino isospin doublet) instead of 4: the reduction from 4 to 2 has to do with the fact that 4*4 complex matrices are 2-fold cover of the real Dirac algebra. More precisely, the real Dirac algebra is isomorphic to 2*2 matrices of quaternions instead of complex numbers, hence there are only 2 columns. A beauty of this matrix spinor (isospin doublet) approach is that you can model internal Lorentz group transformation acting on one side of the spinor as $\exp(\epsilon^{ab}\gamma_a\gamma_b)\psi$ (a,b = 0, 1, 2, 3), and weak group transformation acting on the other side of the same spinor similarly as $\psi \exp(\epsilon'^{ab}\gamma_a\gamma_b)$ (a, b = 1, 2, 3). How about quarks and strong interactions? Hint: you have to go beyond Dirac algebra.

Note that there are two kinds (external and internal) of Lorentz transformations:

  1. External Lorentz transformation (global rotation portion of local diffeomorphism) on differential forms (e.g. electromagnetic gauge field 1-form $A = A_{\mu}dx^{\mu}$) is performed in the spacetime manifold characterized by differential vector base $dx^{\mu}$.
  2. Internal Lorentz transformation $\exp(\epsilon_{ab}\gamma_a\gamma_b)$ on Dirac-algebra-valued objects is performed in the internal spacetime characterized by the whole Dirac algebra (spinor field $\psi$) or bi-vector/vector portion of it (spin connection 1-form $\omega$/tetrad 1-form $e$).

Most of times, we go back/forth (and get away with it) between the external and internal spaces (e.g. Kahler-Dirac fermion) without explicitly mentioning it, thanks to the soldering 1-form vielbein/frame/tetrad $e$, which is a vector in spacetime manifold characterized by differential vector base $dx^{\mu}$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $\gamma_{a}$.

In flat spacetime, the vielbein/frame/tetrad $e$ has the form: $$ e = e^{a}_{\mu}\gamma_adx^{\mu} = \delta^{a}_{\mu}\gamma_adx^{\mu}, $$ which nicely bridges the two spaces, "soldering" $\gamma_a$ and $dx^{\mu}$ via delta function $\delta^{a}_{\mu}$ and producing Minkowskian metric $$g_{\mu\nu} \sim tr(e^{a}_{\mu}\gamma_ae^{b}_{\nu}\gamma_b) = tr(\gamma_\mu \gamma_\nu) \sim \eta_{\mu\nu} . $$

Note that $$ <0|e^{a}_{\mu}|0>= \delta^{a}_{\mu} $$ is a non-zero vacuum expectation value of the 1-form field $e$, which breaks both the external (diffeomorphism) and internal local Lorentz invariances. Without this God-given symmetry breaking effect, our spacetime will be metric-less and devoid of the notion of distance and time interval $$g_{\mu\nu} = 0.$$ An (almost) metric-less spacetime would be perfect playground for wormholes and time/space travelers.

  • 2
    $\begingroup$ I don't understand this question. 1. "differential space manifold" is not a technical term. What do you mean? 2. The spin connection is not a spinor, so it is unclear why you're talking about it being a "vector" or a "bivector" (it is not a "bivector"), and what it has to do with the question of what a spinor is. $\endgroup$
    – ACuriousMind
    Oct 22, 2018 at 18:26
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    $\begingroup$ Comments: 1. Dirac-matrices should multiply from left not right. 2. And no, that's not a Dirac spinor. $\endgroup$
    – Qmechanic
    Oct 22, 2018 at 19:28
  • $\begingroup$ It doesn't matter @Qmechanic. The coefficients like $\psi_s$ and $\psi_v^a$ are just numbers (Grassmann odd though), not columns. $\endgroup$
    – MadMax
    Oct 22, 2018 at 19:31
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    $\begingroup$ A Dirac spinor $\psi$ is a $4\times 1$ column vector. $\endgroup$
    – Qmechanic
    Oct 22, 2018 at 19:33
  • 6
    $\begingroup$ @MadMax Please try to minimize the number of individual edits you make. Each edit should fix all the problems with your question that you see at the time, and you should bunch together individual changes and only make a batch of them at once; if you find yourself editing any given post more than 3-5 times in total, you're probably doing something wrong. This particular post has been edited 40 times, which is way too many, so please don't edit it again, and keep the same principle in mind for other posts you make. $\endgroup$
    – David Z
    Oct 23, 2018 at 5:20

3 Answers 3

  1. There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.

  2. The spin connection is formally a connection form on a $\mathrm{SO}(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $\mathfrak{so}(1,3)$-valued. So as a 1-form it has components $\omega_\mu$ and as a $\mathfrak{so}(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $\gamma$-matrices form a basis of $\mathfrak{so}(1,3)$, so we get your expansion $\omega = \omega_\mu\mathrm{d}x^\mu = {\omega_\mu}^{ab}[\gamma_a, \gamma_b] \mathrm{d}x^\mu$. For more on the spin connection, see also this answer of mine.

  3. A spinor field is now a function that takes values in some spinorial representation of $\mathfrak{so}(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.

However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $\mathfrak{so}(1,3)$-valued field that acts on other fields that take values in some representation of $\mathfrak{so}(1,3)$.

The expression $$ \psi = \psi_s + \psi_v^a \gamma_a +\dots$$ you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $\gamma$-matrices act on the spinor, they are not a basis for its space.

  • $\begingroup$ A spinor lives in the same operator space (4*4 matrices, if you will) spanned by the Dirac vector base $\gamma_a$. The column spinor is just an idempotent projection (left/right ideal) of the matrix spinor. $\endgroup$
    – MadMax
    Oct 22, 2018 at 19:55
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    $\begingroup$ @MadMax That is a non-standard claim. Why do you think so? $\endgroup$
    – ACuriousMind
    Oct 22, 2018 at 19:56
  • $\begingroup$ See the book: books.google.com/… $\endgroup$
    – MadMax
    Oct 22, 2018 at 19:58

These are K{\"a}hler-Dirac fermions

As was pointed out in the comments, consider a $4 \times 1$ column representing a Dirac spinor. Clearly one can embed this in a $4 \times 4$ matrix of zeros and the equation \begin{equation} (\gamma^{\mu}\partial_{\mu} - m)\Psi = 0 \quad \quad (1) \end{equation} makes sense. In fact one can write any $4 \times 4$ matrix as $$ \Psi = f_0 + f_{\mu}\gamma^{\mu} + \frac{1}{2}f_{\mu \nu} \gamma^{\mu} \gamma^{\nu} + \frac{1}{6}f_{\mu \nu \sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma} + f_{0123}\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3} $$ including the column from above.

Now, consider the exterior derivative and its adjoint, $d$ and $\delta$. One can construct a $p$-form field as a linear combination of $p$-forms as $$ \omega = f_{0} + f_{\mu} dx^{\mu} + \frac{1}{2}f_{\mu \nu} dx^{\mu} dx^{\nu} + \frac{1}{6}f_{\mu \nu \sigma}dx^{\mu}dx^{\nu}dx^{\sigma} + f_{0123} dx^{0}dx^{1}dx^{2}dx^{3} $$ and apply the operator $(d - \delta)$ to $\omega$. What one finds is that the action of $\gamma^{\mu}\partial_{\mu}$ on $\Psi$ is the same as the action of $(d - \delta)$ on $\omega$. But since $\Psi$ is a $4 \times 4$ matrix, Eq. (1) is four copies of the Dirac equation, one for each column. In turn, $$ ((d - \delta) - m)\omega = 0 \quad \quad (2) $$ is equivalent to four copies of the Dirac equation since the two operators act identically on their respective objects. However, Eq. (2) is valid for any metric, while Eq. (1) is only good in flat spacetime, and gauging the derivative does not make Eq.(1) and (2) equal.

The references I learned this from are Banks and Rabin.

I hope this helps, let me know if I should clarify, maybe I can add some more later.

  • $\begingroup$ Minor remarks: 1, Your $\omega$ is not a $p$-form, but a formal sum of $p$-forms. 2. That any 4x4 matrix can be spanned by sums of products of $\gamma$-matrices is a manifestation of the fact that the Clifford algebras in even dimension $n$ are the full complex matrix algebras of dimension $2^{n/2}$. 3. I think the information in your comment - that these are Kähler-Dirac fermions and not the standard Dirac fermions - would also fit well in this answer. $\endgroup$
    – ACuriousMind
    Oct 30, 2018 at 12:03
  • $\begingroup$ @ACuriousMind Thanks for the comments. I appreciate you looking over it. I'll make some edits. $\endgroup$
    – kηives
    Oct 30, 2018 at 16:14

I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".

The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $\mathbb R^{1,3}$ itself, and so that for elements $v,w\in \mathbb R^{1,3}$ we have $vw = \langle v,w\rangle$, where $vw$ is the product in the Dirac algebra, and $\langle v,w\rangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4\times 4$ matrices, as you know well.

If we write $\gamma^0,\gamma^1,\gamma^2,\gamma^3$ for the image of an orthonormal basis of $\mathbb R^{1,3}$ in the Dirac algebra, then it is not hard to see that elements of the form $\gamma_{i_1}\cdots\gamma_{i_n}$ indeed generate the Dirac algebra over $\mathbb R$ (or $\mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.

The invertible elements of the Dirac algebra contain an important subgroup, namely $\text{Spin}(\mathbb R^{1,3})$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.

  • $\begingroup$ Nope, "Dirac spinor" spans the whole "Dirac algebra" (it's actually the gist of the question! not just some sub-space of it), while Spin(1,3) is just the bi-vector sub-space where spin connection/Lorentz algebra is valued in. $\endgroup$
    – MadMax
    Oct 22, 2018 at 23:20
  • $\begingroup$ Maybe you mean this: the spinor representation of $\text{Spin}(\mathbb R^{1,3})$ extends to a representation (i.e. a left module) of the Dirac algebra. Since the latter is simple, it actually is a direct sum of copies of this. In that sense, Dirac spinors can be seen as elements of the Dirac algebra. I'm not sure how useful this is: it is a direct sum of 4 copies of them, which in the matrix representation each consist of matrices with a single nonzero column. When taking linear combinations of elements of different copies, you get quantities that only indirectly have to do with Dirac spinors. $\endgroup$
    – doetoe
    Oct 22, 2018 at 23:38
  • $\begingroup$ Actually there are only 2 copies of them: an electron and a neutrino. The reduction from 4 to 2 has to do 4*4 complex matrices are 2-fold cover of real Clifford algebra. $\endgroup$
    – MadMax
    Oct 22, 2018 at 23:42

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