Investigate the conditions under which a group representation defines a central simple division algebra and its relationship to representation conjugacy. | Step-by-Step Solution

Hello! This is a beautiful and deep problem that connects representation theory with the theory of central simple algebras. Let's break it down!

1. What We're Solving:

We want to understand when an absolutely irreducible representation ρ of a finite group G (with character values in field K) gives rise to a central simple K-division algebra, and how to determine its class in the Brauer group.

2. The Approach:

This problem sits at the intersection of several major areas:

• Representation theory to understand how G acts on vector spaces
• Field theory to understand the extension L/K
• Algebra theory to construct division algebras
• Brauer group theory to classify these algebras

The key insight is that group representations naturally create matrix algebras, and under the right conditions, these become division algebras.

3. Step-by-Step Solution:

Step 1: Set up the group algebra Start with the group algebra K[G]. Since ρ is absolutely irreducible with character values in K, we know:

• ρ: G → GL_n(L) for some n
• The character χ_ρ takes values in K
• ρ remains irreducible over any extension of L

Step 2: Construct the endomorphism algebra The key object is End_K[G](V), where V is the representation space. By Schur's lemma:

• Since ρ is absolutely irreducible, End_K[G](V) is a division algebra
• This division algebra is central over its center

Step 3: Determine when we get a K-division algebra The crucial condition is the Schur index. The representation defines a central simple K-division algebra when:

• The Schur index m_K(ρ) > 1
• This happens when ρ cannot be realized over K (even though its character has values in K)

Step 4: Find the division algebra explicitly When the conditions are met, the division algebra D is:

• D ≅ End_K[G](V) where V is viewed as a K[G]-module
• D has dimension m² over K, where m is the Schur index
• The center of D is K

Step 5: Determine the Brauer class The Brauer group class [D] is determined by:

• The cohomological invariants of the representation
• Specifically, it's related to H²(Gal(L/K), L*) when L/K is Galois
• The class depends on how the Galois group acts on the representation

4. The Answer:

The representation ρ defines a central simple K-division algebra if and only if:

• 1. Main condition: The Schur index m_K(ρ) > 1 (i.e., ρ cannot be realized over K)

• 2. The division algebra: D = End_K[G](V) where V is the K[G]-module

• 3. Brauer class: [D] ∈ Br(K) is determined by the 2-cocycle measuring the obstruction to realizing ρ over K

Specifically:

• If ρ can be defined over K, then we get a matrix algebra (Brauer class = 0)
• If ρ requires the extension L/K, then [D] is non-trivial and encodes this "twisting"

5. Memory Tip:

Think of it this way: "Characters live in K, but representations might need L"

When the character has values in K but the representation "insists" on living in the larger field L, that obstruction creates a division algebra. The Brauer class measures exactly how much bigger field you need!

This is a wonderfully rich problem that shows how abstract algebra concepts work together. The beauty is that purely representation-theoretic data (characters, irreducibility) determines purely algebraic data (division algebras, Brauer classes)!