# Download An elementary exposition of Grassmann's Ausdehnungslehre, or by Joseph V. Collins PDF

By Joseph V. Collins

Excerpt from An easy Exposition of Grassmann's Ausdehnungslehre, or idea of Extension

The sum qf any variety of vectors is located via becoming a member of the start element of the second one vector to the tip aspect of the 1st, the start aspect of the 3rd to the top aspect of the second one. and so forth; the vector from the start aspect of the 1st vector to the top element of the final is the sum required.

The sum and distinction of 2 vectors are the diagonals of the parallelogram whose adjoining facets are the given vectors.

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**Extra info for An elementary exposition of Grassmann's Ausdehnungslehre, or Theory of extension**

**Example text**

N = s11 , . . , sk11 , . . , s1n , . . , sknn we have that h( hσ1 , . . , hσn ) = h( s11 , . . , sk11 , . . , s1n , . . , sknn ) and h( a ) = a for a ∈ A I claim that this is the same thing as a group structure on A, with multiplication a · b = h( a, b ). The unit element is given by h( ); the inverse of a ∈ A is h( a−1 ) since h( a, h( a−1 ) ) = h( h( a ), h( a−1 ) ) = h( a, a−1 ) = h( ), the unit element Try to see for yourself how the associativity of the monad and its algebras transforms into associativity of the group law.

N on the alphabet A A−1 , where σ ˜i = σi if σi ∈ T (A), and − σ ˜i = (σi ) if σi ∈ T (A)−1 . Of course we still have to remove possible substrings of the form aa−1 etc. h Now let us look at algebras for the group monad: maps T (A) → A such that for a string of strings α = σ 1 , . . , σn = s11 , . . , sk11 , . . , s1n , . . , sknn we have that h( hσ1 , . . , hσn ) = h( s11 , . . , sk11 , . . , s1n , . . , sknn ) and h( a ) = a for a ∈ A I claim that this is the same thing as a group structure on A, with multiplication a · b = h( a, b ).

Proof. Suppose M : E → C has a limiting cone (C, µ) in C. e. such that νE / GM (E) Dq qq qq q GM (e) νE qq q# GM (E ) e commutes for every E → E in E. 50 ν˜ E This transposes under the adjunction to a family (F D → M E|E ∈ E0 ) and the naturality requirement implies that ν˜E / ME F Dq qq qq qq M (e) ν˜E q# ME commutes in C, in other words, that (F D, ν) is a cone for M in C. There is, therefore, a unique map of cones from (F D, ν˜) to (C, µ). Transposing back again, we get a unique map of cones (D, ν) → (GC, G ◦ µ).