First of all I am a physicist with a decent knowledge of graduate-level geometry. I'm studying Spin Geometry from Bär Lecture Notes and I have some trouble understanding what spinors are from his definition. I'll recap some information here to avoid the needing of reading the lecture notes.

Given a vector space $V$ and a symmetric bilinear form $\beta$ we build the Clifford algebra $Cl(V,\beta)$ with Clifford product $\cdot_C$

Given an algebra homomorphism $\phi: Cl(V,\beta) \rightarrow Cl(V,\beta)$

This splits the algebra as:

$Cl(V,\beta) = Cl^0(V,\beta) \oplus Cl^1(V,\beta)$

With $Cl^0(V,\beta)$ the +1 eigenspace of $\phi$ and $Cl^1(V,\beta)$ the -1 eigenspace of $\phi$

Now i will choose $V = \mathbb{R}^n$ because it's the most straight forward vector space to work with.

Then he defines the Pin and Spin groups as:

$$Pin(n) = \{v_1 \cdot_C ... \cdot_C \; v_m \in Cl(\mathbb{R}^n) | v_i \in S^{n-1} \; , \; m\in \mathbb{N}_0 \}$$


$$Spin(n) = Pin(n)\cap Cl^0(\mathbb{R}^n,\beta) $$

So to my understanding, $Pin(n)$ is the group built multiplying a number $m$ of unitary elements of $\mathbb{R}^n$ while $Spin(n)$ is built only with an even amount of multiplications.

For $n=2m$ he defines 2 quantities:

$$z_j = (e_{2j-1}-ie_{2j})/2 $$ $$\bar{z}_j = (e_{2j-1}+ie_{2j})/2 $$

To me this seems the basis, split into right-handed and left-handed, for the complexified version of $\mathbb{R}^n$ but I'm not sure if it is the case.

Now the questions.

Question 1

He defines:

$$z(j_1, ..., j_k) = z_{j_1} \cdot_C \; ... \cdot_C\; z_{j_k} \cdot_C\; \bar{z}_1 \cdot_C\;...\cdot_C\;\bar{z}_m$$

What does this quantity represent? What is it?

An explanation in geometrical terms would be awesome

Question 2

He defines:

$$\Sigma_n^\pm = span\{z(j_1, ..., j_k) | k=0, ..., m \; j_i \; ordered\}$$ With $+$ for even $m$ and $-$ for odd $m$

He calls elements of such a space spinors, left and right-handed.

How do these elements relate to the "complex vector-type" of spinors usually encountered in physics and to the $Spin(n)$ group above?

Those should be the mathematical counterpart of the physics Weyl spinors, but I can't see how it is possible to recover them.

  • $\begingroup$ I don't know all that much about spinors, but this seems to be related to pure spinors. Notice that $z_i, \bar z_j$ are null vectors with $z_1,\dotsc, z_n$ mutually orthogonal and $\bar z_1,\dotsc,\bar z_n$ mutually orthogonal. So $Z=\bar z_1\dotsb\bar z_n = \bar z_1\wedge\dotsb\wedge\bar z_n$ is a simple $m$-vector (or a blade, or decomposable, or whatever your terminology) in the complexified algebra $$C_{\mathbb C}(V,\beta) := \mathbb C\otimes C(V,\beta) \cong C(C\otimes V,\beta).$$ $\endgroup$ Commented Sep 12, 2023 at 17:46
  • $\begingroup$ Such an $m$-vector natural represents the maximally isotropic subspace $$V_Z = \mathrm{span}\{\bar z_1,\dotsc,\bar z_m\} \subseteq \mathbb C\otimes V,$$ and $Z\bar z_i = 0$ for all $\mathbb z_i$. Thus $\Sigma^\pm_n$ as a right $C_{\mathbb C}(V,\beta)$-module is annihilated by $V_Z$. $\endgroup$ Commented Sep 12, 2023 at 17:47
  • $\begingroup$ Having looked at these notes, I would say that this is a very nonstandard treatment of spinors and spin geometry, at least in how they define the vector space of spinors and Weyl spinors. If you would like a text that deals with spinors rigorously in a way that is readily adaptable to physics, I'd suggest Mathematical Gauge Theory by Hamilton, or Spin Geometry by Michelson and Lawson. $\endgroup$
    – Chris
    Commented Sep 12, 2023 at 20:59

1 Answer 1


I will answer your questions first defining the "classical" complex spinor representation of $Cl(\mathbb R^{2m})$, this is quite natural/geometrical being defined only in terms of wedging and contractions, it is related to complex manifolds and more generally to almost Hermitian manifolds. Then I will prove that this is isomorphic to the representation defined by Bär. This will answer question 2, the answer to question 1 will come up from the proof of the isomorphism.

I usually think of elements of the Clifford algebra as endomorphisms of the complex spinor bundle $S= \oplus_k \Lambda^{k,0}\mathbb R^{2m} $ aka the "complex spinor representation", the latter splits in positive and negative Weyl spinors as $S^+ = \oplus_{\text{k even}} \Lambda^{k,0}\mathbb R^{2m}$ and $S^- = \oplus_{\text{k odd}} \Lambda^{k,0}\mathbb R^{2m}$ the action of $Cl(\mathbb R^{2n})\otimes \mathbb C$ is defined by setting for $\alpha \in \Lambda^{k,0}\mathbb R^{2m}$

$$z_j \cdot \alpha = \sqrt{2} \ z_j \wedge \alpha$$ $$\bar z_j \cdot \alpha = -\sqrt{2} \ \alpha(z^j,\cdot)$$

(Recall that $\Lambda^{k,0} \mathbb R^{2m}$ is the complex vector space generated by $z_{i_1}\wedge \dots \wedge z_{i_k}$ for any subset $\{i_1,\dots, i_k\} \subset \{1,\dots, m\}$ of cardinality $k$.)

This is one of the most geometric ways of thinking about spinors because it ties them to complex geometry and shows that in that setting they are a natural object to consider.

Bär is defining a representation in a different way, instead of taking $S^+$ and $ S^-$ as above he considers his $\Sigma^\pm$ and defines the action by multiplication in the algebra. One possible reason for doing so is that Bär's definition can be given without going through the definition of almost complex structures and forms of type $(p,q)$.

In any case, there is only one irreducible representation of $Cl(\mathbb R^{2n})$ up to isomorphism indeed there is an isomorphism of representations $F: S^\pm\simeq \Sigma^\pm$ defined as follows:

$$F(z_{i_1}\wedge \dots \wedge z_{i_k}) = 2^{-\frac k 2} z(i_1,\dots, i_k)$$

to see that $F$ interwines the action of the Clifford algebra we look at the behaviour under multiplication by $z_j$ and $\overline{z}_j$ (because these elements generate the complexified Clifford algebra $Cl(\mathbb R^{2n})\otimes \mathbb C$), we find that

$$\small{z_j \cdot (z_{i_1}\wedge \dots \wedge z_{i_k}) = 2^{1/2}z_j \wedge z_{i_1}\wedge \dots \wedge z_{i_k}\overset{F}{\mapsto } 2^{-(k+1-1)/2}z(j, i_1,\dots, i_k) = {z}_j \cdot F(z_{i_1}\wedge \dots \wedge z_{i_k})}$$

Similarly, we consider the case $j= i_1 $ for simplicity, the other cases are analogous, then $$\small{\overline z_{i_1} \cdot (z_{i_1}\wedge \dots \wedge z_{i_k}) = - 2^{1/2} \iota_{z^{i_1}}(z_{i_1}\wedge \dots \wedge z_{i_k})\overset{F}{\mapsto } -2^{-(k-2)/2}z(i_2,\dots, i_k) = \overline{z}_{i_1} \cdot F(z_{i_1}\wedge \dots \wedge z_{i_k})}$$

Where we used that $$\overline{z}_{i_1}\cdot z(i_1,\dots, i_k) = (\bar z_{i_1}\cdot z_{i_1})z(i_1,\dots, i_k) = -2 z(i_1,\dots, i_k)$$

which relies on the followiing (painful to derive) identity in $Cl(\mathbb R^{2n})\otimes \mathbb C$: $$(\bar z_{j}\cdot z_{j})\cdot \bar z_j = -\frac 1 2(1 + i e_j \cdot e_{j+1})\cdot\bar z_j = -2 \bar z_j.$$

(this is why the term $\bar z_1\dots \bar z_m$ appears in the definition of $z(i_1,\dots, i_k)$, so that we can use this identity)

This showes that the representation $S^\pm$ and $\Sigma^\pm$ are isomorphic.

A good reference for this is Section 3.4 of Friedrich.Dirac Operators in Riemmanian Geometry

  • $\begingroup$ Yes the product is with a fixed element of dimension $n/2$, i was wondering why is that and how this relates to spinors. Also could you elaborate more about the spinor bundles $S^{\pm}$? Like the range where $k$ runs in and how does elements of an exterior algebra relate to spinors $\endgroup$
    – LolloBoldo
    Commented Sep 12, 2023 at 12:28
  • $\begingroup$ What is the definition of spinor that you take as "primitive"? For me a spinor is defined as an element of $S= S^+\oplus S^-$. $k \in \{0,m\}$, for the definition of forms of type $(p,q)$ you can look up any introductory text on complex geometry. The relation between the two definition is worked out in Friedrich book. $\endgroup$ Commented Sep 12, 2023 at 13:32
  • $\begingroup$ I've always heard of spinors as "complex vectors living in a representation of Spin(n)", but i dont see how to relate such elements $\endgroup$
    – LolloBoldo
    Commented Sep 12, 2023 at 14:31
  • 1
    $\begingroup$ @LolloBoldo Thank you, you are a man of your word! Let me know if you want me to clarify something. $\endgroup$ Commented Sep 18, 2023 at 16:05
  • 1
    $\begingroup$ It is fairly clear for now, i'll let you know if i need other clarifications :) $\endgroup$
    – LolloBoldo
    Commented Sep 18, 2023 at 20:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .