Sum of commuting Lie algebra representations
Let
Then
Proof
Since
π ( [ π₯ , π¦ ] ) = π 1 ( [ π₯ , π¦ ] ) + π 2 ( [ π₯ , π¦ ] ) = π 1 ( π₯ ) π 1 ( π¦ ) β π 1 ( π¦ ) π 1 ( π₯ ) + π 2 ( π₯ ) π 2 ( π¦ ) β π 2 ( π¦ ) π 2 ( π₯ ) = π 1 ( π₯ ) π 1 ( π¦ ) + π 2 ( π₯ ) π 2 ( π¦ ) β π 1 ( π¦ ) π 1 ( π₯ ) β π 2 ( π¦ ) π 2 ( π₯ ) . + π 1 ( π₯ ) π 2 ( π¦ ) + π 2 ( π₯ ) π 1 ( π¦ ) β π 2 ( π¦ ) π 1 ( π₯ ) β π 1 ( π¦ ) π 2 ( π₯ ) = π ( π₯ ) π ( π¦ ) β π ( π¦ ) π ( π₯ ) as required.