fundamental theorem of coalgebras
Fundamental Theorem of Coalgebras. Let (C,Δ,ε) be a coalgebra over a field k and x∈C. Then there exists subcoalgebra D⊆C such that x∈D and dimkD<∞.
Proof. Let
Δ(x)=∑ibi⊗ci. |
Consider the element
Δ2(x)=∑iΔ(bi)⊗ci=∑i,jaj⊗bij⊗ci. |
Note that we may assume that (aj) are linearly independent and so are (ci). Let D be a subspace
spanned by (bij). Of course dimkD<∞. Furthermore x∈D, because
x=∑i,jε(aj)ε(ci)bij. |
We will show that D is a subcoalgebra, i.e. Δ(D)⊆D⊗D. Indeed, note that
∑i,jΔ(aj)⊗bij⊗ci=∑i,jaj⊗Δ(bij)⊗ci |
and since ci are linearly independent we obtain that
∑jΔ(aj)⊗bij=∑jaj⊗Δ(bij) |
for all i. Thus
∑jaj⊗Δ(bij)∈C⊗C⊗D |
and since aj are linearly independent, we obtain that Δ(bij)∈C⊗D for all i,j. Analogously we show that Δ(bij)∈D⊗C, thus
Δ(bij)∈C⊗D∩D⊗C=D⊗D, |
(please, see this entry (http://planetmath.org/TensorProductOfSubspacesOfVectorSpaces) for last equality) which completes the proof. □
Remark. The category of finite dimensional coalgebras is dual to the category of finite dimensional algebras (via dual space
functor), so one could think that generally they are similar
. Unfortunetly Fundamental Theorem of Coalgebras is major diffrence between algebras and coalgebras. For example consider a field k and its polynomial algebra k[X]. Then whenever f∈k[X] is such that deg(f)>0, then a subalgebra generated by f is always infinite dimensional (if deg(f)=0 then subalgebra generated by f is k). This can never occur in coalgebras.
Title | fundamental theorem of coalgebras |
---|---|
Canonical name | FundamentalTheoremOfCoalgebras |
Date of creation | 2013-03-22 18:49:22 |
Last modified on | 2013-03-22 18:49:22 |
Owner | joking (16130) |
Last modified by | joking (16130) |
Numerical id | 6 |
Author | joking (16130) |
Entry type | Theorem |
Classification | msc 16W30 |