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I understand the abstract theoretical notion that Spin(3,1) is a double cover of SO(3,1), but I cannot process this when it comes to my choice of representation.

I am using the following representation of SO(3,1)

$$ R=\exp ( \frac{1}{2}B) $$

where B is a bivector of clifford algebra over $\mathbb{R}^{3,1}$.

Do I have a representation of SO(3,1), or of Spin(3,1).

If I only have SO(3,1), what is missing from my representation to get Spin(3,1)?

Anon21
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  • So how does $R$ act on a vector? And does $-R$ give the same action? – Hans Lundmark Feb 20 '23 at 05:34
  • @HansLundmark $v'=RvR^{-1}=-Rv (-R)^{-1}$ – Anon21 Feb 20 '23 at 11:13
  • So you have two different elements in the Clifford algebra that correspond to the same element of $SO(3,1)$ (namely the linear transformation $v \mapsto v'$), and those elements belong to the double cover $\mathrm{Spin}(3,1)$. – Hans Lundmark Feb 20 '23 at 11:50
  • @HansLundmark So to sum up, and to illustrate the difference, the set of all R represents SO(3,1), and the set of all $\pm R$ represents Spin(3,1)? – Anon21 Feb 20 '23 at 11:51
  • Hmm, that's not how I would think about it... By the way, when you're using the word "represents", do you refer to the technical meaning of a "representation of a group on a vector space", or just the colloquial meaning "things that correspond to one another somehow"? I'm not certain that I understand exactly what you're asking about. – Hans Lundmark Feb 20 '23 at 11:59
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    A bivector exponential is an element of the spin group, not the special-orthogonal group. It is a multivector, not a transformation (or matrix). – mr_e_man Feb 21 '23 at 21:49

1 Answers1

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The bivectors are elements of the Clifford algebra which are linear combinations of those of the form $uv$ with $u$ and $v$ perpendicular (WLOG you could pick $u=e_i,v=e_j$ standard basis vectors with $i<j$). When you exponentiate them, you're plugging them into a power series, and the result is once again an element of the Clifford algebra, more specifically in the spin group which resides inside the algebra.

For example, consider $e_ie_j$. You can verify $R=\exp(\theta e_ie_j)=\cos(\theta)+\sin(\theta)e_ie_j$. If you define the spin group as being generated by products of evenly-many unit or pseudo-unit vectors (those which square to $\pm1$ in the algebra, meaning their (pseudo)-Euclidean squared norm is $\pm1$), we can see this $R$ is in the spin group by writing it as $e_i(-\cos\theta e_i+\sin\theta e_j)$.

The vector space is considered a subspace of the Clifford algebra. (The algebra is graded, starting with scalars at grade $0$, then vectors at grade $1$, then bivectors, etc.) And it is closed under the conjugation-action of the spin group. (One could conversely define the spin group via this property. The inverse of a product of vectors is the reverse product times $\pm1$ depending on the parity of the number of vectors, and this is extended linearly to an anti-automorphism of the algebra which is used in place of inversion for the conjugation action.) Thus, for an element $R$ of the spin group, we can define $\phi(R)$ to be a linear operator acting on the vector space via $\phi(R)v:=RvR^{-1}$.

If we assume the vector space has an ordered basis, all linear operators acting on it may be represented as matrices, which means in turn $\phi(R)$ can be written down as a matrix. This gives a $2$-to-$1$ homomorphism $\phi:\mathrm{Spin}(V)\to\mathrm{SO}(V)$. Since $\phi(-R)=\phi(R)$, the fibers are $\pm R$. In other words, $\pm R$ are distinct elements of the spin group, but turn into the same rotation in the special orthogonal group.

For illustration, in the 3D case, we can pick $\pm 1$ (where $R=1$) to be $0^\circ$ and $360^\circ$ rotations around an axis. In the vectorial world, rotating $360^\circ$ has the same final effect as rotating $0^\circ$. But in the spinorial world, we can think of an element of $\mathrm{Spin}(3)$ as a path in $\mathrm{SO}(3)$ from the identity matrix to another matrix (IOW, an "animation" of rotation), modulo endpoint-preserving homotopy. Within $\mathrm{SO}(3)$, the loop comprised of all rotations around an axis, which goes from the identity matrix back to itself, is noncontractible. However, traversing this loop twice can be visualized as a twice-wound rubber band which actually is contractible within $\mathrm{SO}(3)$, as illustrated by Dirac's belt / plate trick.) So $\pm1$ are different elements of the spin group, but turn into the same rotation.

anon
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  • Do exponentials of bivectors form a complete representation of Spin?, Or are some elements of Spin missing from it? – Anon21 Mar 10 '23 at 18:09
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    In the definite case (i.e. not indefinite signature), yes, since then ${\rm Spin}(n)$ is compact and compact groups are the images of their lie algebras under exponentiation. I expect not otherwise. In particular, ${\rm Spin}(3,1)\cong{\rm SL}_2\Bbb C$ and famously $[\begin{smallmatrix}-1&1\0&-1\end{smallmatrix}]\not\in\exp({\frak sl}_2\Bbb C)$. The range $\exp({\frak g})$ in general is a neighborhood of a Lie group $G$'s identity, generates the rest of $G$ as an abstract group, and IIRC the range contains all but possibly a subset of positive codimension (in particular, measure zero). – anon Mar 10 '23 at 19:58
  • Do you know if the exponentials of the bivectors plus the pseudo-scalars (in 3+1D) maps to ${\rm Spin}^c(3,1)$? Or is it ${\rm Spin}(3,1)\times U(1)$? – Anon21 Mar 10 '23 at 20:56
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    In the complexified Clifford algebra, the exponential of an imaginary scalar plus a bivector is an element of ${\rm Spin}^c(3,1)$. – anon Mar 10 '23 at 22:05
  • Thank you so much. Just two more questions. Do you know what the lie algebra of ${\rm Spin}^c(3,1)$ is? Also, do you know what group the exponential of an imaginary scalar plus a bivector is, in the real Clifford algebra 3+1D? – Anon21 Mar 10 '23 at 22:07
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    The lie algebra of ${\rm Spin}^c(3,1)$ is the direct sum of the imaginary scalars and the bivectors in the Clifford algebra, with the commutator operation. Imaginary scalars are not elements of the real Clifford algebra. – anon Mar 10 '23 at 22:11
  • So, If I were to do $\exp ( \mathfrak{so}(3,1) + \mathfrak{u}(1)) $, I would get ${\rm Spin}(3,1) \times {\rm U}(1)$ (and not ${\rm Spin}^c(3,1)$), correct? – Anon21 Mar 10 '23 at 22:15
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    Depends on your definitions. If you interpret ${\frak so}(3,1)$ as the usual lie algebra of $4\times4$ matrices, and ${\frak u}(1)$ as $i\Bbb R I_4$, and $\exp$ is given by the usual power series, then $\exp({\frak so}(3,1)+i\Bbb R I_4)$ $=$ $S^1{\rm SO}(3,1)$ $\cong$ $S^1\times_{\Bbb Z_2}{\rm SO}(3,1)$, which is neither. If you instead want the space $\Lambda^2\Bbb R^{3,1}$ of bivectors in the real Clifford algebra $C\ell(3,1)$ and $i\Bbb R$ in the complex Clifford algebra (containing the real algebra), then $\exp(\Lambda^2\Bbb R^{3,1}+i\Bbb R)$ is ${\rm Spin}^c(3,1)$. – anon Mar 10 '23 at 22:59
  • I'm sorry I may have made a mistake in my last question. I wanted to ask what group associates with the exponential of bivectors + pseudo-scalars of CL(3,1)? Still I appreciate your answer, it helps me visualize what is going on. Is the exponential of bivectors + pseudo-scalars in CL(3,1) the Spin(3,1) $\times$ U(1) group? – Anon21 Mar 10 '23 at 23:28
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    I define ${\rm Spin}^c(3,1)=S^1{\rm Spin}(3,1)\cong S^1\times_{\Bbb Z_2}{\rm Spin}(3,1)$ as a subgroup of the complex(ified) Clifford algebra, where $S^1\subset\Bbb C^\times$, in which case the exponential of bivectors + pseudoscalars in $C\ell(3,1)$ is not the same copy of ${\rm Spin}^c(3,1)$ but is isomorphic to it. I believe this is not isomorphic, only "approximately"/"locally" isomorphic. – anon Mar 11 '23 at 00:10
  • here, https://en.wikipedia.org/wiki/Spin_structure#SpinC_structures it suggest ${\rm Spin}^c(3,1)$ can be defined as $({\rm Spin}(3,1) \times {\rm U(1)}) / \mathbb{Z}_2$, where $\mathbb{Z}_2$ represents the two coverings of ${\rm Spin}(3.1)$ over SO(3,1). The two coverings corresponds with $v' = RvR^{-1}$ (first covering) and $v' = -R v (-R)^{-1}$ (second covering). Since I take $R=\exp \mathbf{f}$, then I only get one covering. Consequently, it seems the exponential of bivector + pseudo-scalar gives exactly ${\rm Spin}^c(3,1)$. True or false? – Anon21 Mar 12 '23 at 10:28