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QUICK PHRASES: topological space with continuous operations that satisfy homotopy versions of group laws
A H-group is a set equipped with the structure of a topological space and three operations:
|Operation name||Arity of operation||Operation description and notation|
|Multiplication or product||2||A binary operation (infix operator) termed the multiplication or product. The product of and is denoted .|
|Identity element (or neutral element)||0||A 0-ary operation which gives a constant element, denoted by (sometimes also as ), termed the identity element or neutral element.|
|Inverse map||1||A unary operation (superscript operator) termed the inverse map. The inverse of is denoted .|
satisfying two kinds of compatibility conditions -- continuity conditions and homotopy versions of group law conditions.
The continuity conditions are as follows:
|Multiplication or product||2 (so it's a map )||continuous as a map from (equipped with the product topology) to . In other words, the multiplication is jointly continuous.||Joint continuity is strictly stronger than separate continuity, which would mean continuity in each input holding the other input fixed.|
|Identity element||0 (it's a constant element )||no condition. As such, we may impose the condition that the map from a one-point space to sending the point to the identity element is continuous, but this condition is vacuously true.|
|Inverse map||1 (so it's a map )||continuous as a map from to itself with the equipped topology. In other words, is continuous.|
The homotopy version of group law conditions are as follows:
|Condition name||Arity of mappings about which we are making homotopy assertion||Condition description||Comments|
|Homotopy version of associativity||3||Consider the two mappings given by and . These two maps are homotopic to each other as maps from , where the former is equipped with the product topology arising from the topology on .||Note that if the multiplication is associative, then it is homotopy associative, because in that case the two mappings are exactly equal.|
|Homotopy version of identity element (or neutral element)||1||Consider the following three mappings : the identity map , the map , and the map . All of these are homotopic to each other.||Note that if is an identity element in the usual sense of the word, then the three maps are equal and hence homotopic.|
|Homotopy version of inverse element||1||Consider the map given by . This map is homotopic to the constant map .||Note that if is actually a two-sided inverse of for all , the maps are equal identically.|
It's important to note here that unlike the ordinary associativity, identity element, and inverse element conditions, the homotopy versions cannot be checked separately for each tuple of elements. Rather, to check for the truth of the homotopy version, we need to look at the mappings (such as ) in their entirety as we try to homotope them.
If is equipped with a topology that makes it a contractible space, then the homotopy versions of the group laws are vacuously satisfied, because all maps are nullhomotopic. In other words, in this case, we can select any continuous choice for the multiplication, identity element, and inverse map.