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Isn't this a trivial corollary?
Calculus on surfaces and chain ruleConstructing submanifolds. Did I understand this right?Some trivial questions about Tangent SpacesHow to see this is a normal vector field?Finding tangent plane to $2$ dimensional submanifold of $mathbbR^4$Generic condition for vector fields/normal sectionsHow to recover the tangent space from the metricInequality for gradients under different metricsShow that if $f$ is a smooth function, $M$ is a manifold and $x$ is a local extremum of $f$ on $M$, then $D_f(x)(v) = 0$ in the tangent space.Why $dim(ker T_z f)=dim(T_z(f^-1(c)))$?
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Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?
derivatives differential-geometry orthogonality submanifold tangent-spaces
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playdisplaydis
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Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?
derivatives differential-geometry orthogonality submanifold tangent-spaces
asked 8 hours ago
playdisplaydis
394 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?
derivatives differential-geometry orthogonality submanifold tangent-spaces
asked 8 hours ago
playdisplaydis
394 bronze badges
$endgroup$
Let $U subseteq mathbb R^n$ be an open subset and let $M subseteq U$ be a $k$-dimensional submanifold of $mathbb R^n$. Consider a differentiable function $f: U longrightarrow mathbb R$. There's a corollary that states, that if $f big|_M$ takes on a local extremum at a point $p in M$, then the gradient $nabla f(p)$ is normal to $M$ at $p$, i.e. $nabla f(p) perp T_pM$, where $T_pM$ is the tangent space. I've studied and understood the (short) proof, but isn't this statement absolutely trivial or have I got things mixed up? Let me explain: Assuming $p$ is an extremum, it follows that the differential vanishes, i.e. $nabla f(p) = 0$, since this is a necessary condition. But the zero vector is orthogonal to everything. So what's the point of this corollary?
derivatives differential-geometry orthogonality submanifold tangent-spaces
derivatives differential-geometry orthogonality submanifold tangent-spaces
asked 8 hours ago
playdisplaydis
394 bronze badges
asked 8 hours ago
playdisplaydis
394 bronze badges
asked 8 hours ago
playdisplaydis
394 bronze badges
asked 8 hours ago
playdisplaydis
394 bronze badges
asked 8 hours ago
playdisplaydis
394 bronze badges
394 bronze badges
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2 Answers
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Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).
The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
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$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.
Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
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2 Answers
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2 Answers
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$begingroup$
Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).
The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
$endgroup$
add a comment |
$begingroup$
Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).
The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
$endgroup$
add a comment |
$begingroup$
Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).
The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
$endgroup$
Take$$M=(x,y)inmathbb R^2mid x^2+y^2=1$$and let $f(x,y)=x^2+y^2$. Then $f|_M$ is constant, and therefore it has an extreme point at every point of $M$. However, $nabla f$ is never equal to $0$, at any point of $M$ (although, as the corollory states, it is orthogonal to $M$ at each point).
The gradient has to be $0$ indeed, if the point at which we are computing it is both an extreme point of $f$ and an interior point of $M$ (if you see $f$ as a subset of $mathbb R^n$), but points of submanifolds usually are not interior points.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
194k24 gold badges152 silver badges269 bronze badges
194k24 gold badges152 silver badges269 bronze badges
add a comment |
add a comment |
$begingroup$
$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.
Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.
Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
$endgroup$
add a comment |
$begingroup$
$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.
Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
$endgroup$
$p$ is not necessarily a local extremum on $U$, it's a local extremum on $M$.
Consider for instance $f$ defined on $mathbbR^2$ by the square of the euclidean norm; then on the circle $S^1$, it reaches a local extremum at each point, but its gradient is not zero on $S^1$ : $nabla f (p) = 2p$ which is indeed orthogonal to $T_pS^1= p^bot$, but is not $0$
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
answered 8 hours ago
MaxMax
19.4k1 gold badge12 silver badges46 bronze badges
19.4k1 gold badge12 silver badges46 bronze badges
add a comment |
add a comment |
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