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  1. Show that $\mathfrak{sl}_2(\mathbf C)$ is isomorphic to $\mathfrak{o}_3(\mathbf C)$ as lie-algebras.

This paper 1 mention such an isomorphism. My question is "how it is easy to check that the defined map is a lie-algebra isomorphism"? We need to show that is linear (I think it is true since the map is defined on basis elements?) And that it preserves the brackets, which can be checked on basis elements. Can you please tell me how to check the conditions in an accurate way using the map given in the paper?

  1. Include that there is a homomorphism of groups $SL_2(C)\to SO_3(C)$ and find its kernel.

The first part follows from knowing that for any finite dimensional Lie algebra $g$ there is a unique, up to isomorphism, simply connected Lie group and to any Lie algebra homomorphism $\mathfrak g \to \mathfrak h$ uniquely lifts to a Lie group homomorphism $G\to H$. Now, how can we compute the kernel of the group homomorphism? Is the lie-algebra isomorphism involved here?

Many thanks

Lam18373
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    Let $\mathfrak{gl}(V)$ denotes the space of linear maps from $V$ to itself viewed as a Lie algebra via the commutator map, so that a representation of a Lie algebra $\mathfrak g$ is a homomorphism of Lie algebras $\rho\colon \mathfrak g \to \mathfrak{gl}(V)$. For any Lie algebra $\mathfrak g$, you have a natural representation of $\mathfrak g$ on itself, $\mathrm{ad}\colon \mathfrak g \to \mathfrak{gl}(\mathfrak{g})$ given by $\mathrm{ad}(x)(y) = [x,y]$ ($\forall x,y \in \mathfrak g$). For $\mathfrak{sl}_3$ this gives you a map to $\mathfrak{gl}(\mathbb C^3)$. – krm2233 Jun 23 '23 at 10:42
  • Yes, here it is $sl_2(C)$ so it is $sl_2(C)\to gl(sl_2(C)$, where is $o_3$ involved then? – Lam18373 Jun 23 '23 at 10:57
  • If you pick a basis of $\mathfrak{sl}_2(\mathbb C)$, then you can identify $\mathfrak{gl}(\mathfrak{sl}_2(\mathbb C)$ with $\mathfrak{gl}_3(\mathbb C)$. Then you can either a) just try and compute the image (e.g. using your basis) or b) try to see if there is a nondegenerate symmetric bilinear form on $\mathfrak{sl}_2(\mathbb C)$ which is invariant for the action of $\mathfrak{sl}_2(\mathbb C)$. If you can find one, then the image of $\mathrm{ad}$ will have to lie in the orthogonal Lie algebra of that form. – krm2233 Jun 23 '23 at 11:08
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    Sorry, I have just seen the paper you refer to (I missed the link before). If you are given a linear map, to check $\phi\colon \mathfrak g_1 \to \mathfrak g_2$ is a homomorphism of Lie algebras you have to check that $\phi([x,y])= [\phi(x),\phi(y)]$ for all $x,y \in \mathfrak g_1$. Since the Lie bracket is bilinear, it is enough to check this for all $x,y$ in a given basis, so for $\mathfrak{sl}_2(\mathbb C)$ you can use ${E,H,F}$. Since the bracket is alternating, you really only need to check the ordered pairs $(H,E),(H,F)$ and $(E,F)$. – krm2233 Jun 23 '23 at 11:17
  • It is okay, thanks. Can you give an idea for how can I compute the kernel of the lie group homomorphism? – Lam18373 Jun 23 '23 at 11:32
  • See the links in no.7 in https://math.stackexchange.com/q/3972981/96384 for near-duplicates. – Torsten Schoeneberg Jun 23 '23 at 14:48
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    That is, https://math.stackexchange.com/q/276380/96384, https://math.stackexchange.com/q/3866709/96384, https://math.stackexchange.com/q/423419/96384, https://math.stackexchange.com/q/3356528/96384. – Torsten Schoeneberg Jun 23 '23 at 16:29
  • Thanks @Torsten Schoeneberg the problem is that i need to show this without using the killing form or wedge product.. can you give a way for 2? Is there exist some theorem or approach to find the kernel of a lie group homomorphism in such case.. – Lam18373 Jun 23 '23 at 18:21
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    It may be simpler to understand the Lie group homomorphism first, and from that deduce the Lie algebra hom... – paul garrett Jun 24 '23 at 18:52
  • Yes, but I could not find out how to define such a homomorphism @paul garrett – Lam18373 Jun 24 '23 at 19:11

2 Answers2

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There is a substantial listing of homomorphisms such as $SL_2(\mathbb C)\to SO(3,\mathbb C)$, at https://www-users.cse.umn.edu/~garrett/m/v/sporadic_isogenies.pdf Probably more than you really wanted! :) But the point is that there are several such things, and they're (in principle) well-known.

In the case at hand, $SL_2(\mathbb C)$ acts by conjugation on the three-dimensional space of two-by-two complex matrices with trace $0$, and preserves the (non-degenerate) quadratic form $Q(x,y)=\mathrm{trace}(xy)$...

paul garrett
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  • Thanks, I am still miss how we get the group hom to $SO_3(C)$ from what you mention in the paper – Lam18373 Jun 25 '23 at 05:51
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    @Lam18373 They explain that in their answer. $SL_2$ acts by conjugation on $\mathfrak{sl}_2$ which has dimension 3 so we have a homomorphism $SL_2 \to GL(3,\mathbb{C})$ and since it preserves the trace form (or equivalently preserves the Killing form) which is a quadratic form the image is contained in $SO(3,\mathbb{C})$ – Callum Jun 25 '23 at 12:31
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$SL(2;\mathbb{C})$ is the double cover of the special orthogonal group $SO(3;\mathbb{C})$.

This follows partly because:

  1. There is a bijective isometry $$(\mathbb{C}^{3},||\cdot||^2) ~\cong ~ (sl(2;\mathbb{C}),\det) $$ $$\mathbb{C}^{3}~\ni~\vec{x}~=~(x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~\sum_{k=1}^3x^k\sigma_k~\in~ sl(2;\mathbb{C}), $$ $$ ||\vec{x}||^2 ~=~\sum_{k=1}^3x^{k} x^{k} ~=~-\det(\sigma), \tag{1}$$ from the $3$-dimensional complex vector space $(\mathbb{C}^3,||\cdot||^2)$ endowed with the standard bilinear (as opposed to sesquilinear) form to the Lie algebra $$ sl(2;\mathbb{C})~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(\sigma)=0\} ~=~ {\rm span}_{\mathbb{C}} \{\sigma_k \mid k=1,2,3\},\tag{2}$$ endowed with the determinant. Here $\sigma_k$ denote the Pauli matrices.

  2. The adjoint representation ${\rm Ad}: SL(2;\mathbb{C})\times sl(2;\mathbb{C})\times sl(2;\mathbb{C})$ given by $$ g\quad \mapsto\quad {\rm Ad}(g)\sigma~:= ~g\sigma g^{-1}, \qquad g\in SL(2;\mathbb{C}),\qquad\sigma~\in~ sl(2;\mathbb{C}), \tag{3} $$ which is length preserving, i.e. $g$ is an orthogonal transformation. In other words, there is a Lie group homomorphism $${\rm Ad}: SL(2,\mathbb{C}) \quad\to\quad O(sl(2;\mathbb{C});\mathbb{C})~\cong~ O(3;\mathbb{C}) .\tag{4}$$

  3. Since ${\rm Ad}$ is a continuous map and $SL(2,\mathbb{C})$ is a connected set, the image ${\rm Ad}(SL(2,\mathbb{C}))$ must again be a connected set. In fact, one may show that there is a surjective Lie group homomorphism
    $${\rm Ad}: SL(2,\mathbb{C}) \quad\to\quad SO(sl(2;\mathbb{C};\mathbb{C})~\cong~ SO(3;\mathbb{C}) .\tag{5} $$

  4. Let us now return to OP's title question: The kernel of the Lie group homomorphism ${\rm Ad}$ is $${\rm ker}({\rm Ad})~=~\{\pm {\bf 1}_{2 \times 2}\}~\cong~\mathbb{Z}_2 ,\tag{6}$$ due to e.g. Schur's Lemma.

Qmechanic
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  • Notes for later: Alternative construction: $\quad s(2,\mathbb{C}):={ \sigma \in {\rm Mat}{2\times 2}(\mathbb{C}) \mid \sigma^t =\sigma }$ $={\rm span}{\mathbb{C}}{{\bf 1}, i\sigma_1,i\sigma_3}$ with $\det$. $\quad g\mapsto\rho(g)\sigma:=g\sigma g^t$. – Qmechanic Jun 02 '25 at 11:28