Exact sequence giving kernel of mapping from tensor square to exterior square
Let be any group. There is a natural exact sequence as follows:
- denotes the third homology group for trivial group action of on .
- denotes the universal quadratic functor and denotes the abelianization of .
- denotes the tensor square of , i.e., the tensor product of with itself where both parts of the compatible pair of actions are the action of on itself by conjugation.
- is the exterior square of .
The maps are as follows:
- The first map is mysterious (some kind of connecting homomorphism?)
- The second map, , sends PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- The third map, , is the natural quotient map that sends a generator of the form to the corresponding element .
- Perfect implies natural mapping from tensor square to exterior square is isomorphism
- Kernel of natural homomorphism from tensor square to group equals third homotopy group of suspension of classifying space
- Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup