Meaning of terms in a direct sum decomposition of an algebra

In summary, the conversation discusses studying subalgebras of the indefinite orthogonal algebra ##\mathfrak{o}(m,n)## and its decomposition into a direct sum of smaller subalgebras. The notation used is unclear and raises questions about the meaning of certain terms, such as ##\bar{x}## and ##\wedge^2(x)##. The conversation also discusses the relationship between this decomposition and the canonical representation of ##\mathfrak{o}(m,n)## on ##\mathbb{R}^{n+m}##, as well as an embedding between subalgebras. The speaker is seeking further clarification and references on the topic.
  • #1
Rabindranath
10
1
Let's say I want to study subalgebras of the indefinite orthogonal algebra ##\mathfrak{o}(m,n)## (corresponding to the group ##O(m,n)##, with ##m## and ##n## being some positive integers), and am told that it can be decomposed into the direct sum $$\mathfrak{o}(m,n) = \mathfrak{o}(m-x,n-x) \oplus \mathfrak{gl}(x) \oplus \wedge^2(x) \oplus \wedge^2(\bar{x}) \oplus (m+n-2x) \otimes (x \oplus \bar{x})$$ where ##x## is some positive integer ##0<x<m,n##.

I understand the first two terms ##\mathfrak{o}(m-x,n-x) \oplus \mathfrak{gl}(x)## — i.e. a partition into a smaller ##\mathfrak{o}## part and a general linear part ##\mathfrak{gl}(x)## — but I am not so sure how the latter terms should be read/interpreted, especially not the ##(m+n-2x) \otimes (x \oplus \bar{x})## part. What "is", say, a ##9 \otimes (5 \oplus \bar{5})##?

If we represent ##\mathfrak{o}(m,n)## schematically as a matrix $$\left(\begin{array}{ccc|c|c} & & & & \\ & A & & E & F \\ & & & & \\\hline & G & & B & C \\\hline & H & & D & B \end{array}\right)$$ the ##A## block corresponds to the ##\mathfrak{o}(m-x,n-x)## subalgebra and the ##B##s to the ##\mathfrak{gl}(x)## subalgebra. I assume that the wedge terms in the direct sum correspond to the ##C## and ##D## blocks in this matrix, and that the ##(m+n-2x) \otimes (x \oplus \bar{x})## term is somehow associated with the blocks ##E##, ##F##, ##G## and ##H##, since ##(m+n-2x) \times x## is the block dimension of ##E## as well as ##F##.

Any interpretative help, references etc. would be greatly appreciated.
 
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  • #2
Your notation is unclear:

  1. If ##x## is an integer, what does ##\bar x## mean?
  2. The exterior algebra formation operator ##\wedge## applies to vector spaces. So ##\wedge(x)## has no meaning and hence neither does ##\wedge^2(x)##.
  3. ##x\oplus \bar x## has no meaning with ##x## being an integer rather than an algebraic structure.
  4. ##(m+n-2x)## is a number, not an algebraic structure, so we cannot take a direct product of it with anything else.
 
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  • #3
andrewkirk said:
Your notation is unclear:

  1. If ##x## is an integer, what does ##\bar x## mean?
  2. The exterior algebra formation operator ##\wedge## applies to vector spaces. So ##\wedge(x)## has no meaning and hence neither does ##\wedge^2(x)##.
  3. ##x\oplus \bar x## has no meaning with ##x## being an integer rather than an algebraic structure.
  4. ##(m+n-2x)## is a number, not an algebraic structure, so we cannot take a direct product of it with anything else.
Your four points basically sum up my questions perfectly. I assume that the integers in question double as names for algebraic structures, and that e.g. ##8## and ##\bar{8}## would be two different structures corresponding to ##x = 8##. Would this make sense?
 
  • #4
So, as far as I understand, ##\wedge^2(x)## would be the algebra of bivectors ##u \wedge v##, for ##u## and ##v## of dimension ##x##. Does this sound like a meaningful statement?
 
  • #5
Where did you get it from? The matrix decomposition appears to be the right starting point, but it all depends on how the basis vectors are chosen and which part operates how on the others. For e.g. how is the embedding
$$
\mathfrak{o}(m-x,n-x) \subseteq \mathfrak{o}(m,n)
$$
defined? It all looks as if taken from the canonical representation of ##\mathfrak{o}(m,n)## on ##\mathbb{R}^n## or ##\mathbb{R}^m## or ##\mathbb{R}^{n+m}.## You haven't told us which convention was used by writing ##\mathfrak{o}(m,n).## Which is the bilinear form that defines orthogonality?

Your question about ##9\otimes (5\oplus \bar 5)## is legit and I expected it to be answered in your sources. Without additional information, I would read it as ##\mathbb{R}^9\otimes (\mathbb{R}^5 \oplus \mathbb{R}^5).##
 
  • #6
fresh_42 said:
You haven't told us which convention was used by writing ##\mathfrak{o}(m,n).## Which is the bilinear form that defines orthogonality?
Thanks for pointing out the missing information. The basis in question is such that the full ##\mathfrak{o}(m+n-x,m+n-x) \equiv \mathfrak{o}(d,d)## metric representative has the block off-diagonal form
$$\eta = \begin{pmatrix} 0 & I_d \\ I_d & 0 \end{pmatrix}$$
(where the block ##I_d## is the ##d \times d## identity matrix). I guess I should also have written out the corresponding full ##\mathfrak{o}(m,n) \oplus \mathfrak{o}(n-x,m-x)## matrix decomposition, i.e.
$$\left(\begin{array}{ccc|c|c|ccc} & & & & & & & \\ & A & & E & F & & & \\ & & & & & & & \\\hline & G & & B & C & & I & \\\hline & H & & D & B & & J & \\\hline & & & & & & & \\ & & & K & L & & A & \\ & & & & & & & \end{array}\right)$$
Here, as before, the ##A## blocks correspond to the ##\mathfrak{o}(n-x,m-x)## part, the ##B## blocks to the ##\mathfrak{gl}(x)## part, and the whole piece from my original post above corresponds, again, to ##\mathfrak{o}(m,n)##.

So, we have something like an embedding
$$\mathfrak{o}(m,n) \oplus \mathfrak{o}(n-x,m-x) \subseteq \mathfrak{o}(d,d)$$
in which, in turn, we have
$$\mathfrak{o}(n-x,m-x) \oplus \mathfrak{gl}(x) \subseteq \mathfrak{o}(m,n)$$
Now, what I want to understand is the meaning of the direct sum terms ##\wedge^2(x)## and ##(m+n-2x) \otimes (x \oplus \bar{x}) = (d-x) \otimes (x \oplus \bar{x})## in my original post, and how these correspond to, respectively, the ##C## & ##D##, and the ##E##, ##F##, ##G## & ##H## (and also ##I##, ##J##, ##K## & ##L##) blocks in the schematical matrix.

Please let me know if I missed your point.

As for the "symbolic integer" notation, I have only encountered it before in cases like when you write e.g. ##\mathbf{248} \otimes \mathbf{248} = \mathbf{1} \oplus \mathbf{248} \oplus \mathbf{3875} \oplus \mathbf{27000} \oplus \mathbf{30380}## for ##E_8## etc., but it is all a bit esoteric to me. My advisor just looked at a result I had obtained and used the notation in question to express its constituents. I have no other sources re this notation, but I would love if someone had any tips for further reading on the subject.
 
  • #8
fresh_42 said:
Your question about ##9\otimes (5\oplus \bar 5)## is legit and I expected it to be answered in your sources. Without additional information, I would read it as ##\mathbb{R}^9\otimes (\mathbb{R}^5 \oplus \mathbb{R}^5).##
This is the highest weights and decomposing tensor products into irridusibles. For example for ##\mathfrak{so}(3)## you would have seen ##V_{\frac32}\otimes V_1 = V_{\frac52}\oplus V_{\frac32}\oplus V_{\frac12}##. Which with this notation is ##\bf{\frac32}\otimes\bf{1} = \bf{\frac52}\oplus\bf{\frac32}\oplus\bf{\frac12}##.
 
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  • #9
fresh_42 said:
Your question about ##9\otimes (5\oplus \bar 5)## is legit and I expected it to be answered in your sources. Without additional information, I would read it as ##\mathbb{R}^9\otimes (\mathbb{R}^5 \oplus \mathbb{R}^5).##
Or ##\mathbb{C}^9\otimes (\mathbb{C}^5 \oplus \mathbb{C}^5)##.

Rabindranath said:
As for the "symbolic integer" notation, I have only encountered it before in cases like when you write e.g. ##\mathbf{248} \otimes \mathbf{248} = \mathbf{1} \oplus \mathbf{248} \oplus \mathbf{3875} \oplus \mathbf{27000} \oplus \mathbf{30380}## for ##E_8## etc., but it is all a bit esoteric to me. My advisor just looked at a result I had obtained and used the notation in question to express its constituents. I have no other sources re this notation, but I would love if someone had any tips for further reading on the subject.

fresh_42 said:
I'm not familiar with these physical groups and their algebras or their zoo of representations.
A good reference I found for representations is
Fulton-Harris, Representation Theory, Springer GTM 129, 2004
which appears to be the standard book in the field. Here are some lecture notes on the topic
https://www2.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf
I very much like Fulton and Harris, but I would say "a standard reference," not "the standard reference." It certainly does have a lot of examples. From Fulton and Harris: "Another aspect of the book that readers may want to approach in different ways is the profusion of examples. These are put in largely for didactic reasons: we feel that this is the sort of material that can best be understood by gaining some direct hands-on experience with the objects involved. For the most part, however, they do not actually develop new ideas; the reader whose tastes run more to the abstract and general than the concrete and special may skip many
of them without logical consequence. (Of course, such a reader will probably wind up burning this book anyway.)"

martinbn said:
This is the highest weights and decomposing tensor products into irridusibles. For example for ##\mathfrak{so}(3)## you would have seen ##V_{\frac32}\otimes V_1 = V_{\frac52}\oplus V_{\frac32}\oplus V_{\frac12}##. Which with this notation is ##\bf{\frac32}\otimes\bf{1} = \bf{\frac52}\oplus\bf{\frac32}\oplus\bf{\frac12}##.
Yes, this is this decomposition, but I am fairly certain that this is not what this notation means. When there is no ambiguity (and even sometimes when there is ambiguity) physicists often denote representations by the dimension of the representation space. I am not saying that I think that this is good notation, I am just stating what physicists actually use.

Physicists would write your example as ##\bf{4}\otimes\bf{3} = \bf{6}\oplus\bf{4}\oplus\bf{2}##. In the example given by @Rabindranath above, the Lie algebra of ##E_8## is 248-dimensional, i.e., ##\bf{248}## denotes the adjoint representation of the Lie algebra. Note that the dimensions on the LHS and RHS of both examples match: ##4 \times 3 = 6 + 4 + 2 = 16## and ##248 \times 248 = 1 + 248 + 3875 + 27000 + 30380 = 61504##.
 
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1. What is a direct sum decomposition of an algebra?

A direct sum decomposition of an algebra is a way of breaking down a larger algebra into smaller, simpler algebras. It involves expressing the algebra as a direct sum of its subalgebras, which are mutually exclusive and together make up the original algebra.

2. How is a direct sum decomposition useful in algebra?

A direct sum decomposition allows us to study the structure of a larger algebra by breaking it down into smaller, more manageable pieces. It also helps in understanding the properties and relationships between the subalgebras and the larger algebra.

3. What are the terms used in a direct sum decomposition of an algebra?

The terms used in a direct sum decomposition of an algebra include "direct sum", "subalgebra", "mutually exclusive", and "together".

4. Can a direct sum decomposition be applied to any algebra?

Yes, a direct sum decomposition can be applied to any algebra, as long as it can be expressed as a direct sum of its subalgebras. However, not all algebras may have a direct sum decomposition.

5. How is a direct sum decomposition different from other types of algebraic decompositions?

A direct sum decomposition is different from other types of algebraic decompositions, such as direct product or tensor product, because it involves breaking down the algebra into mutually exclusive subalgebras. This means that the subalgebras do not overlap and together they make up the entire algebra.

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