Post History
#2: Post edited
by
Prime Mover
·
2022-06-10T11:20:28Z (over 2 years ago)
added definition of entropic structure
- Context: self-study from Warner's "Modern Algebra (1965): Exercise 16.27.
- >Let $\alpha$ and $\beta$ be endomorphisms of an entropic structure $(S, \odot)$ such that $\alpha \odot \beta$ is the identity automorphism, and let $\otimes$ be the operation on $S$ defined by:
- $$x \otimes y = \alpha (x) \odot \beta (y)$$
- for all $x, y \in S$. Then $(S, \otimes)$ is an entropic idempotent algebraic structure.
- Idempotence is trivial, but I'm having trouble proving entropicness.
- Here is my attempt:
- $(w \otimes x) \otimes (y \otimes z)$
- $= (\alpha (w \otimes x) ) \odot (\beta (y \otimes z) )$
- $= (\alpha (\alpha (w) \odot \beta (x))) \odot (\beta (\alpha (y) \odot \beta (z)))$
- $= (\alpha (\alpha (w) ) \odot \alpha (\beta (x))) \odot (\beta (\alpha (y)) \odot \beta (\beta (z)))$ (justified by homomorphism)
- $= (\alpha (\alpha (w) ) \odot \beta (\alpha (y))) \odot (\alpha (\beta (x)) \odot \beta (\beta (z)))$ (as $\odot$ is entropic)
- $= (\alpha (w) \otimes \alpha (y)) \odot (\beta (x) \otimes \beta (z))$
- If we can say that $\alpha (w) \otimes \alpha (y) = \alpha (w \otimes y)$ then we are home and dry, but I can't see why we can make that leap, as it is far from obvious that $\alpha$ and $\beta$ are endomorphisms on $(S, \otimes)$ in the same way that they are on $(S, \odot)$. Or are they? And why?
- What is it I'm missing here?
- Context: self-study from Warner's "Modern Algebra (1965): Exercise 16.27.
- >Let $\alpha$ and $\beta$ be endomorphisms of an entropic structure $(S, \odot)$ such that $\alpha \odot \beta$ is the identity automorphism, and let $\otimes$ be the operation on $S$ defined by:
- $$x \otimes y = \alpha (x) \odot \beta (y)$$
- for all $x, y \in S$. Then $(S, \otimes)$ is an entropic idempotent algebraic structure.
- An **entropic structure** is an algebraic structure $(S, \circ)$ such that $\circ$ is an **entropic operation**, that is:
- $$\forall a, b, c, d \in S: (a \circ b) \circ (c \circ d) = (a \circ c) \circ (b \circ d)$$
- Idempotence is trivial, but I'm having trouble proving entropicness.
- Here is my attempt:
- $(w \otimes x) \otimes (y \otimes z)$
- $= (\alpha (w \otimes x) ) \odot (\beta (y \otimes z) )$
- $= (\alpha (\alpha (w) \odot \beta (x))) \odot (\beta (\alpha (y) \odot \beta (z)))$
- $= (\alpha (\alpha (w) ) \odot \alpha (\beta (x))) \odot (\beta (\alpha (y)) \odot \beta (\beta (z)))$ (justified by homomorphism)
- $= (\alpha (\alpha (w) ) \odot \beta (\alpha (y))) \odot (\alpha (\beta (x)) \odot \beta (\beta (z)))$ (as $\odot$ is entropic)
- $= (\alpha (w) \otimes \alpha (y)) \odot (\beta (x) \otimes \beta (z))$
- If we can say that $\alpha (w) \otimes \alpha (y) = \alpha (w \otimes y)$ then we are home and dry, but I can't see why we can make that leap, as it is far from obvious that $\alpha$ and $\beta$ are endomorphisms on $(S, \otimes)$ in the same way that they are on $(S, \odot)$. Or are they? And why?
- What is it I'm missing here?
#1: Initial revision
by
Prime Mover
·
2022-06-10T10:07:24Z (over 2 years ago)
Endomorphisms on an entropic structure whose pointwise product is the identity automorphism - entropic idempotent structure?
Context: self-study from Warner's "Modern Algebra (1965): Exercise 16.27. >Let $\alpha$ and $\beta$ be endomorphisms of an entropic structure $(S, \odot)$ such that $\alpha \odot \beta$ is the identity automorphism, and let $\otimes$ be the operation on $S$ defined by: $$x \otimes y = \alpha (x) \odot \beta (y)$$ for all $x, y \in S$. Then $(S, \otimes)$ is an entropic idempotent algebraic structure. Idempotence is trivial, but I'm having trouble proving entropicness. Here is my attempt: $(w \otimes x) \otimes (y \otimes z)$ $= (\alpha (w \otimes x) ) \odot (\beta (y \otimes z) )$ $= (\alpha (\alpha (w) \odot \beta (x))) \odot (\beta (\alpha (y) \odot \beta (z)))$ $= (\alpha (\alpha (w) ) \odot \alpha (\beta (x))) \odot (\beta (\alpha (y)) \odot \beta (\beta (z)))$ (justified by homomorphism) $= (\alpha (\alpha (w) ) \odot \beta (\alpha (y))) \odot (\alpha (\beta (x)) \odot \beta (\beta (z)))$ (as $\odot$ is entropic) $= (\alpha (w) \otimes \alpha (y)) \odot (\beta (x) \otimes \beta (z))$ If we can say that $\alpha (w) \otimes \alpha (y) = \alpha (w \otimes y)$ then we are home and dry, but I can't see why we can make that leap, as it is far from obvious that $\alpha$ and $\beta$ are endomorphisms on $(S, \otimes)$ in the same way that they are on $(S, \odot)$. Or are they? And why? What is it I'm missing here?