Mixing
General concept of 'quotient'
Working with group theory I've found multiple times the idea of quotient group as $G/H = {gH | gin G, H < G}$. Nevertheless, you can find similar things in vectorial spaces as $mathbb{R}^2/L = {Pvec{v} = vec{v}^{perp} | vec{v} = vec{v}^L + vec{v}^{perp}, vec{v}in mathbb{R}^2, vec{v}^L in L, vec{v}^{perp} in L^{perp}}$. Here, $L$ is an straight line in $mathbb{R}^2$
Is there a way to understand any quotient of this type without given the proper definition of the quotient (as I did above) nor equivalent relation? A general way to deduce what the quotient means by the meaning of the numerator and denominator?
group-theory vector-spaces quotient-spaces quotient-group
asked Nov 20 '18 at 0:28
Vicky
1437
add a comment |
Working with group theory I've found multiple times the idea of quotient group as $G/H = {gH | gin G, H < G}$. Nevertheless, you can find similar things in vectorial spaces as $mathbb{R}^2/L = {Pvec{v} = vec{v}^{perp} | vec{v} = vec{v}^L + vec{v}^{perp}, vec{v}in mathbb{R}^2, vec{v}^L in L, vec{v}^{perp} in L^{perp}}$. Here, $L$ is an straight line in $mathbb{R}^2$
Is there a way to understand any quotient of this type without given the proper definition of the quotient (as I did above) nor equivalent relation? A general way to deduce what the quotient means by the meaning of the numerator and denominator?
group-theory vector-spaces quotient-spaces quotient-group
asked Nov 20 '18 at 0:28
Vicky
1437
1
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
1
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15
add a comment |
Working with group theory I've found multiple times the idea of quotient group as $G/H = {gH | gin G, H < G}$. Nevertheless, you can find similar things in vectorial spaces as $mathbb{R}^2/L = {Pvec{v} = vec{v}^{perp} | vec{v} = vec{v}^L + vec{v}^{perp}, vec{v}in mathbb{R}^2, vec{v}^L in L, vec{v}^{perp} in L^{perp}}$. Here, $L$ is an straight line in $mathbb{R}^2$
Is there a way to understand any quotient of this type without given the proper definition of the quotient (as I did above) nor equivalent relation? A general way to deduce what the quotient means by the meaning of the numerator and denominator?
group-theory vector-spaces quotient-spaces quotient-group
asked Nov 20 '18 at 0:28
Vicky
1437
Working with group theory I've found multiple times the idea of quotient group as $G/H = {gH | gin G, H < G}$. Nevertheless, you can find similar things in vectorial spaces as $mathbb{R}^2/L = {Pvec{v} = vec{v}^{perp} | vec{v} = vec{v}^L + vec{v}^{perp}, vec{v}in mathbb{R}^2, vec{v}^L in L, vec{v}^{perp} in L^{perp}}$. Here, $L$ is an straight line in $mathbb{R}^2$
Is there a way to understand any quotient of this type without given the proper definition of the quotient (as I did above) nor equivalent relation? A general way to deduce what the quotient means by the meaning of the numerator and denominator?
group-theory vector-spaces quotient-spaces quotient-group
group-theory vector-spaces quotient-spaces quotient-group
asked Nov 20 '18 at 0:28
Vicky
1437
asked Nov 20 '18 at 0:28
Vicky
1437
asked Nov 20 '18 at 0:28
Vicky
1437
asked Nov 20 '18 at 0:28
Vicky
1437
asked Nov 20 '18 at 0:28
Vicky
1437
1437
1
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
1
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15
add a comment |
1
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
1
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15
1
1
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
1
1
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15
add a comment |
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1
You can think of this in the framework of category theory, I suppose.
– T. Bongers
Nov 20 '18 at 0:33
I don't get it. I've seen that $mathbb{R}/mathbb{Z} simeq S^1$ and I don't see how to make the conection
– Vicky
Nov 20 '18 at 0:50
1
Please see my earlier answer on something similar giving some intuitive idea: math.stackexchange.com/questions/988522/…
– P Vanchinathan
Nov 20 '18 at 0:57
Ok, then following your answer in that post, if I wanted to compute $mathbb{Q}/mathbb{Z}$, this would be ${0, 1/2}$ because I wouldn't be able to distinguish between two eleements in $mathbb{Q}$ that differ by integers, true? And $mathbb{R}/mathbb{Z}$ would be $[0, 1]$, i.e., all real numbers between $0$ and one, taking them into account, true?
– Vicky
Nov 20 '18 at 1:10
Therefore, $mathbb{R}/mathbb{Z} notsimeq S^1$ but $mathbb{R}/(2pimathbb{Z}) simeq S^1$, isn't it?
– Vicky
Nov 20 '18 at 1:15