Is it possible to show that so(4) is equal to su(2) direct sum su(2)?

In summary: I was after!Thanks for all your help guys, it was much appreciated, and sorry for the slow reply - I'm an extremely slow worker! :)In summary, the conversation discusses the topic of Lie algebras, specifically focusing on the relationship between so(4) and su(2). The poster is struggling to show that so(4) is equal to the direct sum of two copies of su(2). After receiving some hints, the poster is able to construct a basis for so(4) and show that it is isomorphic to the direct sum of two copies of su(2).
  • #1
Anthony
83
0
Hi guys - long time reader first time poster!
I'm currently getting to grips with the topic of Lie Algebras, and I've come across something that's baffled me somewhat. I've been asked to show:

[tex]so(4) = su(2) \oplus su(2)[/tex]

Where the lower so(n) denotes the Lie Algebra of SO(n) etc. Now, in a previous question, I was asked to show:

[tex]u(2) = su(2) \oplus \mathbb{R} [/tex]

Where [tex]\mathbb{R}[/tex] denotes the set of constant (real) multiples of the matrices of the form [tex]i\mathbb{I}_2[/tex]. This was easy enough; I showed that for each [tex]v\in u(2)[/tex] there exists a [tex]x \in su(2)[/tex] and a [tex]y \in \mathbb{R}[/tex] such that v=x+y, and also that if [tex]A \in su(2) \cap \mathbb{R} [/tex] then A=0.

However, in this new case, I assume the [tex]\oplus[/tex] means the the matrix direct sum, but surely if this is the case, then it's false - since the matrices in so(4) take the form:

[tex]\left(\begin{array}{cccc}0&a&b&c\\-a&0&d&e\\-b&-d&0&f\\-c&-e&-f&0\end{array}\right)[/tex]

Which is not of the form of a matrix direct sum. If anyone could give me any hints as to where my confusion lies, I'd be very greatful - although if you could keep the hints sufficiently vague, as to not to do all the work for me!
 
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  • #2
Actually, I should add. My only (perhaps?) contructive thought was the existence of a certain isomorphism which takes su(2) to a subset of so(4). For a general [tex]v \in su(2) [/tex] we have:

[tex] v = \left(\begin{array}{cc}ia&ic+b\\ic-b&-ia\end{array}\right) = \left(\begin{array}{cc}0&b\\-b&0\end{array}\right) + i\left(\begin{array}{cc}a&c\\c&-a\end{array}\right) = A + iB[/tex]

Then there is an isomorphism G such that:

[tex] G(v) = \left(\begin{array}{cc}A&-B\\B&A\end{array}\right) [/tex]

Since we have:

[tex] G(v_1 v_2) = \left(\begin{array}{cc}A_1 A_2 - B_1 B_2&-(A_1 B_2 + B_1 A_2 )\\A_1 B_2 + B_1 A_2&A_1 A_2 - B_1 B_2\end{array}\right) = \left(\begin{array}{cc}A_1&-B_1\\B_1&A_1\end{array}\right)\left(\begin{array}{cc}A_2&-B_2\\B_2&A_2\end{array}\right) = G(v_1)G(v_2) [/tex]

And so:

[tex]G([v_1,v_2]) = [G(v_1),G(v_2)] [/tex]

But this doesn't give me all of so(4).
 
Last edited:
  • #3
Are you working with real or complex Lie algebras?

It is true that for complex Lie algebras,

[tex]
so_{4} \mathbb{C} \cong su_{2} \mathbb{C} \oplus su_{2} \mathbb{C},
[/tex]

but I don't think that a similar relationship holds for the real Lie algebras so(4) and su(2).

The set of matrices

[tex]
\left\{\left(\begin{array}{cc}i&0\\0&-i\end{array}\right), \left(\begin{array}{cc}0&i\\i&0\end{array}\right), \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\right\}
[/tex]

is a basis for both the real Lie algebra su(2) and the complex Lie algebra [tex]su_{2} \mathbb{C}[/tex]. In the former, all real linear combinations of basis elements are allowed; in the latter, all complex linear combinations of basis elements are allowed.

If you would like, I will give some broad hints in a future post.

Regards,
George
 
Last edited:
  • #4
Anthony said:
However, in this new case, I assume the [tex]\oplus[/tex] means the the matrix direct sum, but surely if this is the case, then it's false - since the matrices in so(4) take the form:
[tex]\left(\begin{array}{cccc}0&a&b&c\\-a&0&d&e\\-b&-d&0&f\\-c&-e&-f&0\end{array}\right)[/tex]
Which is not of the form of a matrix direct sum. If anyone could give me any hints as to where my confusion lies, I'd be very greatful - although if you could keep the hints sufficiently vague, as to not to do all the work for me!


Have you considered that there might be more than one basis you can think of?
 
  • #5
the latex for setting lie algebras is \mathfrak{text}, though i don't know if we support it here:
[tex]\mathfrak{sl}_2[/tex]

incidentally, sl_2 and so_3 are the same lie algebra, but if you pick the standard basis they dont' look it.
 
  • #6
George Jones said:
Are you working with real or complex Lie algebras?
Apologies George - I should have said, the Lie Algebras are complex. Any hints (however vague) would be much appreciated. :)
matt grime said:
Have you considered that there might be more than one basis you can think of?
Hmmm, ok - so at the moment I'm considering a basis of [tex]\mathfrak{so}(4)[/tex] of the form:

[tex]\left{\left(\begin{array}{cccc}0&1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right), \left(\begin{array}{cccc}0&0&1&0\\0&0&0&0\\-1&0&0&0\\0&0&0&0\end{array}\right), \quad \textrm{etc.}\right}[/tex]

But perhaps if I think about suitable combinations of these, I'll get something more like the form I'm looking for?

The help is much appreciated guys. :)
 
Last edited:
  • #7
matt grime said:
incidentally, sl_2 and so_3 are the same lie algebra

In more detail: the real Lie sl(2,C) has a complex structure. When this is taken into account, sl(2,C) with complex structure is isomorphic to the complexification of the real lie algebra so_3.

This is why angular momentum theory in physics quantum theory texts looks like the representation theory of sl_2 given in math texts.

Regards,
George
 
  • #8
Well, I thought I'd get back to you guys and let you know it's all sorted. Here's how I went about it:

1) Contruct a basis for [tex]\mathfrak{so}_4[/tex] with the following matrices:


[tex]\left\{ \left{\left(\begin{array}{cccc}
0&1&0&0\\
-1&0&0&0\\
0&0&0&1\\
0&0&-1&0\end{array}\right),
\left(\begin{array}{cccc}
0&0&0&1\\
0&0&-1&0\\
0&1&0&0\\
-1&0&0&0\end{array}\right),
\left{\left(\begin{array}{cccc}
0&0&1&0\\
0&0&0&-1\\
-1&0&0&0\\
0&1&0&0\end{array}\right),
\left(\begin{array}{cccc}
0&0&-1&0\\
0&0&0&-1\\
1&0&0&0\\
0&1&0&0\end{array}\right),
\left{\left(\begin{array}{cccc}
0&0&0&1\\
0&0&-1&0\\
0&1&0&0\\
-1&0&0&0\end{array}\right),
\left(\begin{array}{cccc}
0&-1&0&0\\
1&0&0&0\\
0&0&0&1\\
0&0&-1&0\end{array}\right) \right\} [/tex]

Which I'll refer to as [tex]\left\{ a_1, a_2, a_3, b_1, b_2, b_3 \right\}[/tex]. Then upon computing the Lie brackets, we see:

[tex] [a_i, a_j] = \epsilon_{ijk} a_k \qquad [b_i, b_j] = \epsilon_{ijk} b_k [/tex]

And so [tex]\left\{a_i\right\} \cong \mathfrak{su}_2[/tex] and also [tex]\left\{b_i\right\} \cong \mathfrak{su}_2[/tex]. Since [tex]\mathfrak{so}_4={a_i} \oplus {b_i}[/tex] we have:

[tex]\mathfrak{so}_4 \cong \mathfrak{su}_2 \oplus \mathfrak{su}_2 [/tex]

Which I think is ok.
 
Last edited:
  • #9
a_2 = b_2. Is this just a typo?

Regards,
George
 
  • #10
Sure is (too many matrix elements!). It ([tex]a_2[/tex]) should be:

[tex]\left(\begin{array}{cccc}
0&0&0&1\\
0&0&1&0\\
0&-1&0&0\\
-1&0&0&0\end{array}\right)[/tex]
 

1. What is decomposition of Lie algebras?

Decomposition of Lie algebras is a process in which a given Lie algebra is decomposed into simpler subalgebras. This is done by identifying certain subspaces within the algebra that have special properties, such as being invariant under the action of the algebra's elements.

2. Why is decomposition of Lie algebras important?

Decomposition of Lie algebras allows us to break down a complex algebra into smaller, more manageable parts. This can help us understand the structure and properties of the original algebra better, and can also provide insight into the relationship between different Lie algebras.

3. What are some common methods for decomposing Lie algebras?

There are several methods for decomposing Lie algebras, including the Cartan decomposition, the Levi decomposition, and the Jordan decomposition. Each of these methods has its own advantages and is useful for different types of Lie algebras.

4. How is decomposition of Lie algebras related to representation theory?

Decomposition of Lie algebras is closely related to representation theory, which studies how algebraic structures, such as Lie algebras, act on vector spaces. By decomposing a Lie algebra into smaller subalgebras, we can obtain a better understanding of its representations and how they are related to each other.

5. Can decomposition of Lie algebras be applied in other areas of mathematics?

Yes, decomposition of Lie algebras has applications in various areas of mathematics, including differential geometry, mathematical physics, and algebraic topology. It can also be used in practical applications, such as in the study of symmetries in physical systems.

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