# Why is Lorentz Group in 3D SL(2, R)?

1. Jul 31, 2014

### maverick280857

Hi,

While reading "Superspace: One Thousand and One Lessons in Supersymmetry" by Gates et al. I came across the following paragraph:

Maybe I haven't understood what exactly they're trying to say here, but

1. Why is the Lorentz Group SL(2, R) instead of SL(2, C)?
2. Why is the two-component spinor real? (Well I guess this follows from question 1).

Any response will be much appreciated!

PS - The book is available freely from http://arxiv.org/abs/hep-th/0108200.

Thanks!

EDIT: I understand that the proper orthochronous Lorentz group $L_{+}^{\uparrow}$ is homomorphic to SL(2, C), i.e. for any $M \in$ SL(2, C), there exists a Lorentz matrix

$$\Lambda = \lambda(M) \in L_{+}^{\uparrow}$$

such that $\Lambda(M_1) \Lambda(M_2) = \Lambda(M_1 M_2)$

and $\Lambda^{-1}(M) = \Lambda(M^{-1})$

But how does going from 4 to 3 dimensions take us from SL(2, C) to SL(2,R)?

Last edited: Jul 31, 2014
2. Jul 31, 2014

### Physics Monkey

I suppose the starting point is to count parameters. In 4d you have 3 rotations and 3 boosts while in 3d you have 1 rotation and 2 boosts.

But perhaps this is too elementary a response?

To say a little more, the gamma matrices may be taken to be multiples of the Pauli matrices, e.g. $\gamma^0 = i Z$, $\gamma^1 = X$, $\gamma^2 = Y$. Play around with different choices to find the reality properties you're after.

Last edited: Jul 31, 2014
3. Aug 2, 2014

### lpetrich

The 3D Lorentz group is SO(2,1), much as the 4D one is SO(3,1).

If you have some computer-algebra software, see if you can find the commutation tables for the Lie algebras for SO(2,1) and SL(2,R). You should try Pauli-like generators of SL(2,R). They should match, to within some overall factor.

It's much like doing them for SO(3) and SU(2), where each axis's rotation generator matches a Pauli matrix.

Spinor - Wikipedia and Spin representation - Wikipedia have some helpful info, but they can get very arcane. But in the latter one, one can find some rules for the reality of spinor representations.
• R = real
• H = Hamilton = quaternions = pseudoreal
• C = complex
Real and pseudoreal reps are equivalent to their complex conjugates, though pseudoreal ones can't be turned into a manifestly real form, as real ones can.

For SO(p,q), here's the reality as a function of (p - q) mod 8:
0: R+R, 1: R, 2: C, 3: H, 4: H+H, 5: H, 6: C, 7: R
For 2 and 6, the spinor reps are complex conjugates of each other.

One gets from this that the SO(2,1) spinor is real, unlike the SO(3) spinor, which is pseudoreal. SL(2,R) is also real.

4. Aug 2, 2014

### maverick280857

Thanks Physics Monkey and Ipetrich.

I have a few questions:

1. What is $x^{\sigma\tau}$ in terms of $x^{\mu}$? Am I correct in thinking that this is just a vector described by a symmetric rank-2 spinor (as mentioned on page 2)

2. What is the basis for defining the derivatives as above?

Last edited: Aug 2, 2014
5. Aug 2, 2014

### haushofer

Isn't this just the fact that in three dimensions spinorreps are two-dim. and one can choose Majorana spinors in three dimensions? (See e.g. Van Proeyen's Tools for SUSY)

6. Aug 3, 2014

### haushofer

Maybe i'm mixing up reps with the particular matrix groups themselves.

7. Aug 3, 2014

### Fredrik

Staff Emeritus
The analysis of the 3+1-dimensional case starts with the observation that
$$(x^0,x^1,x^2,x^3)\mapsto x^\mu\sigma_\mu=\begin{pmatrix}x^0+x^3 & x^1-ix^2\\ x^1+ix^2 & x^0-x^3\end{pmatrix}$$ is a bijection from $\mathbb R^4$ to the real vector space of complex self-adjoint 2×2 matrices. ($\sigma_0$ is the identity matrix. The other three basis vectors are the Pauli spin matrices).

If we use the +--- signature, and use the notation $x$ both for the 4×1 matrix and the corresponding 2×2 matrix, we have $\det x=x^T\eta x$. This implies that every linear operator on this set of matrices that preserves determinants defines a Lorentz transformation. Transformations of the form $u\mapsto uxu^*$ where u is an element of $SL(2,\mathbb C)$ are linear and preserve both the self-adjointness and the determinant.

Now what happens if you instead start with the following map?
$$(x^0,x^1,x^2)\mapsto x^\mu\sigma_\mu=\begin{pmatrix}x^0 & x^1-ix^2\\ x^1+ix^2 & x^0\end{pmatrix}$$ Now the space of matrices appear to be the subset of the set of self-adjoint matrices that have the same number on the upper left as on the lower right. What conditions do you have to impose on the matrix u for the transformation $x\mapsto uxu^*$ to preserve that property? I haven't actually checked, but I'm guessing that $u$ has to have real components, since that's what you need.

Last edited: Aug 3, 2014
8. Aug 4, 2014

### lpetrich

Fredrik, you've demonstrated that the Lorentz group SO(3,1) ~ SL(2,C)

Set $x^0 = 0$ and one can demonstrate that SO(3) ~ SU(2)

Set $x^2 = 0$ and one can demonstrate that SO(2,1) ~ SL(2,R)

One has to set that one to zero to make the right-hand-side matrix all-real.

9. Aug 4, 2014

### TrickyDicky

So for $x^0 = 0$ does it hold PSL(2,C) ~ SL(2,C)?

10. Aug 4, 2014

### lpetrich

The group SL(2,C) is the double cover of the group PSL(2,C), just as the group SU(2) is the double cover of the group SO(3). In other words, elements R and -R of the first group can be identified with an element of the second group.

11. Aug 5, 2014

### samalkhaiat

Proving the group homomorphism $SO( 2 , 1) \cong SL( 2 , \mathbb{ R } )$ is quite complicated and requires good understanding of the conformal group. And, since I’ve just received a complain about my “sophisticated” replies on PF, I will show you how to prove the Lie algebra isomorphism $so( 2 , 1) \cong sl( 2 , \mathbb{ R } )$ which is quite easy, and then I will improve on that by considering the transformations of the two groups.

The Lie algebra of $SO( 2 , 1)$ is
$$[ M_{ a b } , M_{ c d } ] = \eta_{ b c } M_{ a d } - \eta_{ b d } M_{ a c } + \eta_{ a d } M_{ b c } - \eta_{ a c } M_{ b d } , \ \ (1)$$
with $a , b = 0 , 1 , 2$ and $\eta = \mbox{ diag } ( -1 , +1 , -1 )$.

Now, we introduce the following combinations of generators
$$P = M_{ 0 2 } - M_{ 0 1 } , \ \ K = M_{ 0 2 } + M_{ 0 1 } ,$$
and rename the remaining one as
$$D = M_{ 2 1 } .$$
It is now easy to show that, in terms of the generators $P , K$ and $D$, the Lie algebra (1) becomes
$$[ D , P ] = - 2 P , \ \ \ [ D , K ] = 2 K , \ \ \ [ P , K ] = D . \ \ (2)$$
To show that the algebra of $SL( 2 , \mathbb{ R } )$ is exactly of the form (2), recall that every group element $g \in SL( 2 , \mathbb{ R } )$ can be parameterized as
$$g( \epsilon_{ 1 } , \epsilon_{ 2 } , \epsilon_{ 3 } ) = \left( \begin{array}{rr} 1 + \epsilon_{ 1 } & \epsilon_{ 2 } \\ \epsilon_{ 3 } & \frac{ 1 + \epsilon_{ 2 } \epsilon_{ 3 } }{ 1 + \epsilon_{ 1 } } \end{array} \right) . \ \ (3)$$
Close to the identity element, i.e., to first order in the infinitesimal parameters $\epsilon_{ i }$, (3) becomes
$$g( \epsilon_{ 1 } , \epsilon_{ 2 } , \epsilon_{ 3 } ) = \left( \begin{array}{rr} 1 + \epsilon_{ 1 } & \epsilon_{ 2 } \\ \epsilon_{ 3 } & 1 - \epsilon_{ 1 } \end{array} \right) .$$
Using this, the infinitesimal generators can be deduced from
$$X_{ i } = \frac{ \partial g }{ \partial \epsilon_{ i } } |_{ ( 0 , 0 , 0 ) } .$$
They are
$$X_{ 1 } = \left( \begin{array}{rr} 1 & 0 \\ 0 & - 1 \\ \end{array} \right) , \ X_{ 2 } = \left( \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right) , \ X_{ 3 } = \left( \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right) .$$
Thus, by simple matrix multiplications we find the algebra
$$[ X_{ 1 } , X_{ 2 } ] = 2 X_{ 2 } , \ \ [ X_{ 1 } , X_{ 3 } ] = - 2 X_{ 3 } , \ \ [ X_{ 2 } , X_{ 3 } ] = X_{ 1 } .$$
This is exactly the Lie algebra (2) of the group $SO( 2 , 1)$. This means that we have the Lie algebra isomorphism $so( 2 , 1) \cong sl( 2 , \mathbb{ R } )$.

Notice that the algebra (2) has the following differential realizations on functions of single real variable $t \in \mathbb{ R }$,
$$P = \frac{ d }{ d t } , \ \ K = t^{ 2 } \frac{ d }{ d t } , \ \ D = t \frac{ d }{ d t } . \ \ (4)$$
In fact, these are the generators of the conformal group $C( 1 , 0 )$ in $(1 + 0)$ dimensional space-time: $D$ (the generator of the non-compact subgroup $SO(1,1) \subset SO(2,1)$) generates scale transformation on the coordinate $t$, $P$ (the Hamiltonian) generates time translation, and $K$ generates conformal boosts.
So, on the level of algebras, we have the following isomorphism
$$SO( 2 , 1) \cong C( 1 , 0 ) \cong SL( 2 , \mathbb{ R } ) .$$
This actually a special case of an isomorphism between the Lorentz group $SO( 2 , n )$ in $(n + 2)$ dimensions and the conformal group in $n$ dimensions $C( 1 , n - 1 )$. For more details on this and other aspects of the conformal group $C( 1 , n - 1 )$, see (in particular post#6)

So, under $C( 1 , 0 )$ or (which is the same thing) $SO(2,1)$, the coordinate $t$ transforms (infinitesimally) according to (see Eq(1.12) in the link)
$$C( 1 , 0 ) : \ t \rightarrow \bar{ t } = t + \beta + 2 \alpha t - c t^{ 2 } . \ \ (5)$$
Finally, we will now show that the group $SL( 2 , \mathbb{ R } )$ act on $t$ in exactly the same way $C( 1 , 0 )$ does, i.e., starting from $SL( 2 , \mathbb{ R } )$ transformations and arriving at (5). This sounds strange, how can a $2 \times 2$ matrix act on a single variable $t$? This example will show you that this can actually be done.

$SL( 2 , \mathbb{ R } )$ acts naturally on 2-dimensional real vector space. Therefore, under an infinitesimal $SL( 2 , \mathbb{ R } )$ transformations, a 2-component vector transforms according to
$$\left( \begin{array}{C} \bar{ V }_{ 1 } \\ \bar{ V }_{ 2 } \end{array} \right) = \left( \begin{array}{rr} 1 + \alpha & \beta \\ c & 1 - \alpha \end{array} \right) \ \left( \begin{array}{C} V_{ 1 } \\ V_{ 2 } \end{array} \right) , \ \ (6)$$
where $\alpha , \beta$ and $c$ are all real infinitesimal parameters, i.e., $\alpha^{ 2 } = c \beta \approx 0$. This ensures that the determinant of the transformation matrix is one, as it should be for the group $SL( 2 , \mathbb{ R } )$.

Now, if we define $t = V_{ 1 } / V_{ 2 }$, and divide the two equations in (6), we arrive at
$$t \rightarrow \frac{ ( 1 + \alpha ) t + \beta }{ ct + 1 - \alpha } \approx t + \beta + 2 \alpha t - c t^{ 2 } ,$$
which is eq(5) again. Of course, this is not a rigorous proof of the homomorphism $SO( 2 , 1) \cong SL( 2 , \mathbb{ R } )$, but close enough for PF.

Sam

12. Aug 6, 2014

### Korybut

Lorentz group in three dimensions is SO(2,1) and it is NOT isomorphic to SL(2,R). SL(2,R) - is spin group in three dimensions with the signature mentioned above.

SO(2,1)=SL(2,R)/Z_2

Majorana spinors are real because they are Majorana))) Definition of Majorana spinor

13. Aug 7, 2014

### samalkhaiat

What is the meaning of the equal sign in here? And what do we say about groups having the same algebra?

14. Aug 7, 2014

### Korybut

Equal sign means that they are isomorphic.

Equal algebras doesn't mean that group are isomorphic.

O(3) and SO(3) have the same Lie algebra, but they are not isomorphic. Exponential map from algebra to group gives only simply connencted part. One can not build smooth curve from matrices with det=1 to matrices with det=-1.

15. Aug 7, 2014

### samalkhaiat

"Equal algebras" means that the corresponding groups are LOCALLY isomorphic. When I wrote $SO(2,1) \cong SL(2 , \mathbb{ R })$, it should be clear to you that $SL(2 , \mathbb{ R })$ is quotient by its center which happened to coinside with the group $Z_{2}$

16. Aug 8, 2014

### lpetrich

Here's what a projective group is for a matrix group M for n*n matrices. Consider variables ${x_1, \dots x_{n-1}}$. An element of the group is
$$x_i \to \frac{\sum_{j=1}^{n-1}D_{ij}x_j + D_{in}}{\sum_{j=1}^{n-1}D_{nj}x_j + D_{nn}}$$
where D is an element of M.

Thus, Proj(M) = quotient group of M and {all multiples of the n*n identity matrix in M}.

Thus, PSL(2,X) = SL(2,X) / Z2: {I,-I}
for X = R, C, H where H is the quaternions ("Hamilton numbers")
Quaternions can be realized as $q_0 I + i(q \cdot \sigma)$ for a 4-vector of real q's and Pauli matrices σ. T

Likewise, PSU(2) = SU(2) / Z2: {I,-I}

One could proceed further to the octonions, but they have the problem of being non-associative.

Last edited: Aug 8, 2014
17. Aug 8, 2014

### lpetrich

I've tried to catalog the small-Lie-group automorphisms, at least for (semisimple)*(single-element-algebra) ones

GL = general linear, SL = special linear (GL with det = 1)
GL(n,R) = GL(1,R+) * SL(n,R)
GL(n,C) = GL(1,C) * SL(n,C)

U = unitary, SU = special unitary (U with det = 1)
U(n) = U(1) * SU(n)
GL(n,R) and SL(n,R) are analytic continuations of U(n) and SU(n), meaning that one can carry over the SU(n) representation theory. So they both have a fundamental n-D representation with a vector as its basis set and other representations with tensors with varying symmetries.

One can define a more general version of unitary matrices: U(n1,n2) = U(1) * SU(n1,n2)

Their elements D are defined using a Hermitian metric g with signature n1 +'s and n2 -'s:
$$D^\dagger g D = g$$

Orthogonal groups may be defined likewise.
For O(n), DT.D = I
For O(n1,n2), DT.g.D = g

where g is a symmetric real matrix. I'll mainly consider real D.

O(n) and O(n1,n2) consist of disconnected parts. SO(n) (det = 1) is the part of O(n) that contains the identity. SO(n1,n2) if defined by det = 1 is two disconnected parts of O(n1,n2). One has to specialize further to get the O(n1,n2) part with the identity: SO+(n1,n2). Adding to the confusion, that part is sometimes called plain SO(n1,n2)

Thus, O(n) / SO(n) = Z2 (det) and O(n1,n2) / SO(n1,n2) (connected) = Z2 * Z2

For real elements, SO(n1,n2) can be obtained as an analytic continuation of SO(n), but for complex elements, they are equivalent.

Thus the full Lorentz group O(3,1) has the restricted Lorentz group SO(3,1) connected to the identity and being built from rotations and boosts, and 3 other parts related by space reflection, time reflection, and both reflection.

18. Aug 8, 2014

### lpetrich

There's a variant of the orthogonal groups, the symplectic groups, that are constructed with a metric in the same way, but with an antisymmetric metric:
$$g = \left\{ \begin{array}{cc} 0 & I \\ -I & 0 \end{array} \right\}$$
This group has even-dimensional matrices.

Here is a collection of isomorphisms that I've assembled from various sources. The reason why (orthogonal group) = projective of (other groups) is that the orthogonal groups are more-or-less the squares of the other groups or the products of the other groups' two parts. That makes (other group element) and (- other group element) equivalent.

GL(n,R) = GL(1,R+) * SL(n,R)
GL(n,C) = GL(1,C) * SL(n,C)

GL(1,H) = GL(1,R+) * SL(1,H) with quaternions expanded using Pauli matrices

U(n) = U(1) * SU(n)

Thus, one can take the representation theory of SU(n) and carry it over to SL(n,R).

SO(2) family:
SO(2) = U(1)
SO(1,1) = GL(1,R+)

SO(3) family:
SO(3) = PSU(2), SU(2) = Sp(2) = SL(1,H) with expanded quaternions
SO(2,1) = PSU(1,1), SU(1,1) = SL(2,R)

SO(4) family:
SO(4) = P(SU(2) * SU(2))
SO(3,1) = PSL(2,C)
SO(2,2) = P(SU(1,1) * SU(1,1))

SO(5) family:
SO(5) = PSp(4)

SO(6) family:
SO(6) = PSU(4)
SO(5,1) = PSL(2,H) with expanded quaternions
SO(4,2) = PSU(2,2)
SO(3,3) = PSU(3,1)

In SO(3) to SO(6), the projections are dividing out Z2: {I, -I}

19. Aug 8, 2014

### lpetrich

Here are outlines of proofs that I've found. I've found them for the SO(2), SO(3), and SO(4) families using the group elements, but not for the SO(5) or SO(6) families. I have written a Mathematica notebook that does those proofs by construction.

SO(2): $$D = \left\{ \begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right\}$$
Eigenvalues: $e^{i\theta}, e^{-i\theta}$ -- U(1)
Eigenvectors constant

SO(1,1): $$D = \left\{ \begin{array}{cc} \cosh\beta & \sinh\beta \\ \sinh\beta & \cosh\beta \end{array} \right\}$$
Eigenvalues: $e^{\beta}, e^{-\beta}$ -- GL(1,R+)
Eigenvectors constant

SO(3):

For the SU(2) matrices, use $D(g) = g_0 I + i ({\vec g} \cdot \sigma)$ for Pauli matrices σ and unit 4-vector g: |g| = 1. Then, for 3-vector x and SO(3) matrix R(g), $$(R(g) \cdot x) \cdot \sigma = D(g) \cdot (x \cdot \sigma) \cdot D(g)^{-1}$$ Note that R(-g) = R(g).

SO(2,1):

For SU(1,1), change all the Pauli matrices: $(\sigma_1, \sigma_2, \sigma_3) \to (i\sigma_1, i\sigma_2, \sigma_3)$. Likewise for SL(2,R), change all the Pauli matrices: $(\sigma_1, \sigma_2, \sigma_3) \to (i\sigma_1, \sigma_2, i\sigma_3)$. The parameter vector g will no longer be a unit vector when det(D(g)) = 1. But the expression for the SO(3) matrices carries over into SO(2,1).

SO(4):

Define a 4-vector of Pauli matrices: $\sigma_I = \{ I, i\sigma_1,i \sigma_2, i\sigma_3 \}$ Then the SU(2) matrix D(g) = g.σI. For 4-vector x, parameter vectors g1 and g2, and SO(4) matrix R(g1,g2), we get $$(R(g_1,g_2) \cdot x) \cdot \sigma_I = D(g_1) \cdot (x \cdot \sigma_I) \cdot D(g_2)^{-1}$$ Note that R(-g1,-g2) = R(g1,g2).

SO(2,2):

Similar, but with all the Pauli matrices changed as appropriate for SU(1,1) or SL(2,R).

SO(3,1):

Define a different 4-vector of Pauli matrices: $\sigma_R = \{ I, \sigma_1, \sigma_2, \sigma_3 \}$ Also define SL(2,C) matrices D(g) as SU(2) ones with parameter vector g complex instead of real. Then for SO(3,1) matrix R(g), $$(R(g) \cdot x) \cdot \sigma_R = D(g) \cdot (x \cdot \sigma_R) \cdot D(g^*)^{-1}$$ Note that R(-g) = R(g).

20. Aug 19, 2014

### lpetrich

Oops, SO(3,3) = PSU(3,1) = PSL(4,R) also; SU(3,1) = SL(4,R)

But back to the OP, that SO(2,1) Lorentz group was for a toy supersymmetric model in 3D space-time: 2 space, 1 time. The authors started out with that because it's easier to work with than the 4D version, while illustrating the basic ideas of supersymmetry.

21. Oct 16, 2014

### lpetrich

To set the record straight, I've fixed some errors in the low-order-algebra identifications; details are in Explicit construction of isomorphisms of low-rank Lie algebras.. Summary:

SO(2) ~ U(1)
SO(1,1) ~ GL(1,R+)
GL(1,C) ~ GL(1,R+) * U(1)

SO(3) ~ SU(2) ~ SL(1,H) ~ Sp(2)
SO(2,1) ~ SU(1,1) ~ SL(2,R) ~ Sp(2,R)

SO(4) ~ SU(2)*SU(2) , SO(3,1) ~ SL(2,C) , SO(2,2) ~ SU(1,1)*SU(1,1) , SO(2,H) ~ SU(2)*SU(1,1)

SO(5) ~ Sp(4) , SO(4,1) ~ Sp(2,2) , SO(3,2) ~ Sp(4,R)

SO(6) ~ SU(4) , SO(5,1) ~ SL(2,H) , SO(4,2) ~ SU(2,2) , SO(3,3) ~ SL(4,R) , SO(3,H) ~ SU(3,1)

There is an interesting curiosity about the spinor-rep reality of the transverse degrees of freedom for a massless particle. For n+1 space dimensions and 1 time dimension, the transverse degrees of freedom live in n space dimensions. The other two are taken up by the momentum vector. The reality of the spinor rep for the whole space time is the same as that for the transverse dimensions.

Spin(n+1,1) has the same reality as Spin(n) though it may be extrapolated.

1+1D space-time (string sheet): real (0)
2+1D space-time (toy problem here): real (1)
3+1D space-time (our space-time): split into two complex conjugates (2)
9+1D space-time (superstrings): split into two real ones (8)