Definition of the ring of differentiable maps $C^infty(M)$, where $M$ is a differentiable manifold
$begingroup$
Let $M$ be a differentiable manifold. We define the set $C^{infty}(M) := {f : M to mathbb{R} text{ differentiable}}$.
We can endow this set with a ring structure considering the common sum of functions:
$$(f+g)(p) := f(p)+g(p)$$
and the product (instead of the composition):
$$(fg)(p) := f(p)g(p)$$
So, the neutral element for $+$ is the constant map zero, and the neutral element for the product is the constant map $1$ or the identity map? (then is a domain?)
Or we consider $(C^{infty}(M),+,circ)$ with $circ$ the composition of differentiable maps?
Thanks.
differential-geometry smooth-manifolds
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
$endgroup$
add a comment |
$begingroup$
Let $M$ be a differentiable manifold. We define the set $C^{infty}(M) := {f : M to mathbb{R} text{ differentiable}}$.
We can endow this set with a ring structure considering the common sum of functions:
$$(f+g)(p) := f(p)+g(p)$$
and the product (instead of the composition):
$$(fg)(p) := f(p)g(p)$$
So, the neutral element for $+$ is the constant map zero, and the neutral element for the product is the constant map $1$ or the identity map? (then is a domain?)
Or we consider $(C^{infty}(M),+,circ)$ with $circ$ the composition of differentiable maps?
Thanks.
differential-geometry smooth-manifolds
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
$endgroup$
1
$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
1
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19
add a comment |
$begingroup$
Let $M$ be a differentiable manifold. We define the set $C^{infty}(M) := {f : M to mathbb{R} text{ differentiable}}$.
We can endow this set with a ring structure considering the common sum of functions:
$$(f+g)(p) := f(p)+g(p)$$
and the product (instead of the composition):
$$(fg)(p) := f(p)g(p)$$
So, the neutral element for $+$ is the constant map zero, and the neutral element for the product is the constant map $1$ or the identity map? (then is a domain?)
Or we consider $(C^{infty}(M),+,circ)$ with $circ$ the composition of differentiable maps?
Thanks.
differential-geometry smooth-manifolds
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
$endgroup$
Let $M$ be a differentiable manifold. We define the set $C^{infty}(M) := {f : M to mathbb{R} text{ differentiable}}$.
We can endow this set with a ring structure considering the common sum of functions:
$$(f+g)(p) := f(p)+g(p)$$
and the product (instead of the composition):
$$(fg)(p) := f(p)g(p)$$
So, the neutral element for $+$ is the constant map zero, and the neutral element for the product is the constant map $1$ or the identity map? (then is a domain?)
Or we consider $(C^{infty}(M),+,circ)$ with $circ$ the composition of differentiable maps?
Thanks.
differential-geometry smooth-manifolds
differential-geometry smooth-manifolds
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
edited Nov 28 '18 at 13:49
Brahadeesh
6,15742361
6,15742361
asked Nov 28 '18 at 12:48
user540275user540275
717
asked Nov 28 '18 at 12:48
user540275user540275
717
asked Nov 28 '18 at 12:48
user540275user540275
717
717
1
$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
1
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19
add a comment |
1
$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
1
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19
1
1
$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
1
1
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $mathbb{R}$. So, the multiplicative identity is the constant map $1$.
Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = mathbb{R}$. So, composition cannot in general define a multiplication on $C^infty(M)$.
The question
(then is a domain?)
does not make sense to me. Are you perhaps asking if $C^infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
$endgroup$
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
add a comment |
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$begingroup$
$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $mathbb{R}$. So, the multiplicative identity is the constant map $1$.
Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = mathbb{R}$. So, composition cannot in general define a multiplication on $C^infty(M)$.
The question
(then is a domain?)
does not make sense to me. Are you perhaps asking if $C^infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
$endgroup$
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
add a comment |
$begingroup$
$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $mathbb{R}$. So, the multiplicative identity is the constant map $1$.
Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = mathbb{R}$. So, composition cannot in general define a multiplication on $C^infty(M)$.
The question
(then is a domain?)
does not make sense to me. Are you perhaps asking if $C^infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
$endgroup$
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
add a comment |
$begingroup$
$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $mathbb{R}$. So, the multiplicative identity is the constant map $1$.
Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = mathbb{R}$. So, composition cannot in general define a multiplication on $C^infty(M)$.
The question
(then is a domain?)
does not make sense to me. Are you perhaps asking if $C^infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
$endgroup$
$M$ is an arbitrary differentiable manifold, so the "identity map" does not make sense unless $M$ itself is $mathbb{R}$. So, the multiplicative identity is the constant map $1$.
Again, since $M$ is an arbitrary differentiable manifold, composition does not necessarily make sense. It only makes sense if $M = mathbb{R}$. So, composition cannot in general define a multiplication on $C^infty(M)$.
The question
(then is a domain?)
does not make sense to me. Are you perhaps asking if $C^infty(M)$ is an integral domain? If so, the answer is no, and it is a good exercise to find two nonzero smooth functions on $M$ whose product is the zero function.
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
answered Nov 28 '18 at 13:47
BrahadeeshBrahadeesh
6,15742361
6,15742361
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
add a comment |
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
It´s not integral domain because for $f,g$ defined as $f(x)=0, x <0$ and $f(x)=sin(x),x leq 0$; $g(x)=sin(x),x < 0$ and $g(x)=0, x geq 0$ we get $f(x)g(x)=0$ but $f,g$ are non-zero functions, right?
$endgroup$
– user540275
Nov 28 '18 at 16:26
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
$begingroup$
@user540275 There are two problems with your example. The first is that it is specialised to the case $M = mathbb{R}$. It would be better to produce an example for the case of an arbitrary differentiable manifold rather than only for $mathbb{R}$. Secondly, the functions $f$ and $g$ are not differentiable at $0$, they are only continuous at $0$. But the idea is more or less accurate. Try $f(x) = e^{-x^{-2}}$ for $x > 0$ and $f(x) = 0$ for $x leq 0$, and $g(x)$ appropriately as you had done. Then try to see if the same example generalises to an arbitrary manifold $M$.
$endgroup$
– Brahadeesh
Nov 28 '18 at 16:45
add a comment |
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$begingroup$
How do you compose two maps from $M to mathbb R$? Both maps act on elements of $M$, not $mathbb R$. So composition is out of the question. The first notion is therefore the correct one.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 28 '18 at 12:52
1
$begingroup$
This is basically formalising the idea that the pointwise sum and product of any two $C^infty$ functions is again a $C^infty$ function.
$endgroup$
– gandalf61
Nov 28 '18 at 13:19