Timeline for Group action of $\text{SL}(2, \mathbb{C})$
Current License: CC BY-SA 4.0
33 events
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| Nov 28, 2023 at 14:44 | history | edited | user515818 | CC BY-SA 4.0 |
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| Nov 1, 2023 at 5:41 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:43 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:39 | vote | accept | CommunityBot | ||
| Oct 31, 2023 at 21:14 | comment | added | GH from MO | Comments merged into an answer. | |
| Oct 31, 2023 at 21:10 | answer | added | GH from MO | timeline score: 3 | |
| Oct 31, 2023 at 21:05 | comment | added | user515818 | @GHfromMO It would be fantastic if you include these in an answer. | |
| Oct 31, 2023 at 21:04 | comment | added | Vít Tuček | @GHfromMO Thanks! | |
| Oct 31, 2023 at 20:57 | comment | added | GH from MO | @VítTuček The space $\text{H}^{+}_3$ can be naturally identified with $\mathrm{SL}_2(\mathbb{C})/\mathrm{SU}_2(\mathbb{C})$; this identifaction comes from the left action of $\mathrm{SL}_2(\mathbb{C})$ on $\text{H}^{+}_3$. Hence a natural measure on $\text{H}^{+}_3$ is induced by a right Haar measure on $\mathrm{SL}_2(\mathbb{C})$. However, this right Haar measure is also a left Haar measure, whence the natural measure on $\text{H}^{+}_3$ is in fact invariant under the left action of $\mathrm{SL}_2(\mathbb{C})$. So this left action induces a unitary action on $\mathcal{L}^2(\text{H}^{+}_3)$. | |
| Oct 31, 2023 at 20:45 | comment | added | Vít Tuček | @GHfromMO How does unitarity follow from the existence of two-sided Haar measure? | |
| Oct 31, 2023 at 20:26 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 19:05 | comment | added | GH from MO | Both the left and the right action are unitary, because $\mathrm{SL}_2(\mathbb{C})$ is a unimodular group, hence it has a two-sided Haar measure. The Lie-algebra action is as follows. An element $m\in\mathfrak{sl}_2(\mathbb{C})$ maps the function $P\mapsto f(P)$ to the function $P\mapsto\frac{\partial}{\partial t}f(e^{-t m}P)\mid_{t=0}$. Of course this action is only defined for special functions $P\mapsto f(P)$, e.g. smooth and compactly supported functions are fine. | |
| Oct 31, 2023 at 16:27 | comment | added | GH from MO | These group actions (on functions $\text{H}^{+}_3\to\mathbb{C}$) give rise to Lie-algebra actions (on smooth functions $\text{H}^{+}_3\to\mathbb{C}$) in the usual way. Note that smooth and compactly supported functions are dense in $\mathcal{L}^2(\text{H}^{+}_3)$. | |
| Oct 31, 2023 at 16:24 | comment | added | GH from MO | The usual left-action of $\mathrm{SL}_2(\mathbb{C})$ on $\text{H}^{+}_3$ is defined by quaternionic right-division: $MP:=(\alpha P+\beta)(\gamma P+\delta)^{-1}$. Here $P=z+wj$ is a Hamiltonian quaternion with $k$-component zero, and so is $MP$. This left-action induces a right-action on functions $\text{H}^{+}_3\to\mathbb{C}$: the image of the function $P\mapsto f(P)$ is $P\mapsto f(M P)$. It also induces a left-action on functions $\text{H}^{+}_3\to\mathbb{C}$: the image of the function $P\mapsto f(P)$ is $P\mapsto f(M^{-1} P)$. I continue in the next comment. | |
| Oct 31, 2023 at 16:20 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 16:11 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 15:41 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 12:08 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 12:03 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 11:58 | comment | added | Vincent | I.e. when writing things out in actual quaternions, how do we know no $k$'s pop up? | |
| Oct 31, 2023 at 11:57 | comment | added | Vincent | Wait, something else seems to be wrong. Suppose $\gamma = 0, \delta = 1$. How do we guarantee that $w^*$ is still real when $\alpha$ isn't? | |
| Oct 31, 2023 at 11:56 | comment | added | Vincent | (The right action would look more natural if action is written from the right $((z + wj)B)A = (z + wj)(BA)$. Right now I cannot easily see if it is a left or right action, but left seems more plausible) | |
| Oct 31, 2023 at 11:54 | comment | added | Vincent | Dave Benson is rigth. If the orginal action satisfies $(A(B(z + wj)) = (AB)(z + wj)$ (left action) then the action on functions should be defined by $(Mf)(z + wj) = f(M^{-1}(z + wj))$. If the original action satisfies $(A(B(z + wj)) = (BA)(z + wj)$ (right action) then the action on functions can be $(Mf)(z + wj) = f(M(z+wj))$ | |
| Oct 31, 2023 at 11:51 | comment | added | user515818 | @Vincent For sure! It seems straightforward to verify that the map is, in fact, a group action. | |
| Oct 31, 2023 at 11:49 | comment | added | Dave Benson | In (4), wouldn't you need an inverse somewhere? Otherwise composition gets reversed. | |
| Oct 31, 2023 at 11:49 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 11:48 | comment | added | Vincent | Thanks for the clarification in the post! So the question boils down to if the first group action is actually a group action (i.e. respects the matrix multiplication, i.e. satisfies $M(N(z + wj)) = (MN)(z + wj)$) despite the non-commutative nature of $j$. This sounds like something you could verify with pen and paper. | |
| Oct 31, 2023 at 11:41 | comment | added | user515818 | @Vincent Yes, it's the quaternion $j$ that satisfies $zj = j\bar{z}$ for $\mathbb{C}$. | |
| Oct 31, 2023 at 11:41 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 11:35 | comment | added | Vincent | Generally speaking though: if a group $G$ acts on a measure space $X$ by a sufficiently nice action then it also acts on the space $L^2(X)$ in the way you describe | |
| Oct 31, 2023 at 11:33 | comment | added | Vincent | I'm sorry but can you explain your notation a bit more? What is $j$? Is it the quaternion $j$ satisfying $zj = j \overline{z}$ for $z \in \mathbb{C}$? The elements $z + wj$ with $z, w \in \mathbb{Z}$ seem to me to live in a real four-dimensional vectorspace, rather than three dimensional hyperbolic space | |
| Oct 31, 2023 at 9:38 | history | edited | user515818 | CC BY-SA 4.0 |
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| Oct 31, 2023 at 9:25 | history | asked | user515818 | CC BY-SA 4.0 |