Why are there so many different notations for the inner product?
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$$acdot b$$
$$a^Tb$$
$$langle a,brangle$$
$$langle amid brangle$$
$$I(a,b)$$
$$g(a,b)$$
$$a^flat(b)$$
$$alpha(b)$$
Matrices are a different type of object from vectors, so the transpose operation deserves a separate symbol $^T$.
An abstract vector space may have several different inner products, which must have distinct names, so $g$ is reasonable.
But all the others seem superfluous compared to the dot. The notation $acdot b$ correctly suggests bilinearity, that it acts like multiplication. It can also be used in an abstract (no inner product) space for the action of a covector: $alpha(b)=alphacdot b$
vector-spaces notation inner-product-space
asked Sep 10 at 18:31
mr_e_man
1,061422
add a comment |Â
up vote
1
down vote
favorite
$$acdot b$$
$$a^Tb$$
$$langle a,brangle$$
$$langle amid brangle$$
$$I(a,b)$$
$$g(a,b)$$
$$a^flat(b)$$
$$alpha(b)$$
Matrices are a different type of object from vectors, so the transpose operation deserves a separate symbol $^T$.
An abstract vector space may have several different inner products, which must have distinct names, so $g$ is reasonable.
But all the others seem superfluous compared to the dot. The notation $acdot b$ correctly suggests bilinearity, that it acts like multiplication. It can also be used in an abstract (no inner product) space for the action of a covector: $alpha(b)=alphacdot b$
vector-spaces notation inner-product-space
asked Sep 10 at 18:31
mr_e_man
1,061422
The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$acdot b$$
$$a^Tb$$
$$langle a,brangle$$
$$langle amid brangle$$
$$I(a,b)$$
$$g(a,b)$$
$$a^flat(b)$$
$$alpha(b)$$
Matrices are a different type of object from vectors, so the transpose operation deserves a separate symbol $^T$.
An abstract vector space may have several different inner products, which must have distinct names, so $g$ is reasonable.
But all the others seem superfluous compared to the dot. The notation $acdot b$ correctly suggests bilinearity, that it acts like multiplication. It can also be used in an abstract (no inner product) space for the action of a covector: $alpha(b)=alphacdot b$
vector-spaces notation inner-product-space
asked Sep 10 at 18:31
mr_e_man
1,061422
$$acdot b$$
$$a^Tb$$
$$langle a,brangle$$
$$langle amid brangle$$
$$I(a,b)$$
$$g(a,b)$$
$$a^flat(b)$$
$$alpha(b)$$
Matrices are a different type of object from vectors, so the transpose operation deserves a separate symbol $^T$.
An abstract vector space may have several different inner products, which must have distinct names, so $g$ is reasonable.
But all the others seem superfluous compared to the dot. The notation $acdot b$ correctly suggests bilinearity, that it acts like multiplication. It can also be used in an abstract (no inner product) space for the action of a covector: $alpha(b)=alphacdot b$
vector-spaces notation inner-product-space
vector-spaces notation inner-product-space
asked Sep 10 at 18:31
mr_e_man
1,061422
asked Sep 10 at 18:31
mr_e_man
1,061422
edited Sep 10 at 21:34
asked Sep 10 at 18:31
mr_e_man
1,061422
asked Sep 10 at 18:31
mr_e_man
1,061422
asked Sep 10 at 18:31
mr_e_man
1,061422
1,061422
The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20
add a comment |Â
The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20
The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20
add a comment |Â
1 Answer
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up vote
0
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As you said, the notation $acdot b$ suggests bilinearity. The notations $left<a,bright>$ or $left<amid bright>$ or $left(a,bright)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example,
$$left<u,alpha vright> = alpha left<u,vright>, quad left<alpha u,vright> = bar alpha left<u,vright>,
$$
where $bar alpha$ is the complex conjugate of $alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).
answered Sep 10 at 19:27
Kusma
3,440219
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As you said, the notation $acdot b$ suggests bilinearity. The notations $left<a,bright>$ or $left<amid bright>$ or $left(a,bright)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example,
$$left<u,alpha vright> = alpha left<u,vright>, quad left<alpha u,vright> = bar alpha left<u,vright>,
$$
where $bar alpha$ is the complex conjugate of $alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).
answered Sep 10 at 19:27
Kusma
3,440219
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
add a comment |Â
up vote
0
down vote
As you said, the notation $acdot b$ suggests bilinearity. The notations $left<a,bright>$ or $left<amid bright>$ or $left(a,bright)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example,
$$left<u,alpha vright> = alpha left<u,vright>, quad left<alpha u,vright> = bar alpha left<u,vright>,
$$
where $bar alpha$ is the complex conjugate of $alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).
answered Sep 10 at 19:27
Kusma
3,440219
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As you said, the notation $acdot b$ suggests bilinearity. The notations $left<a,bright>$ or $left<amid bright>$ or $left(a,bright)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example,
$$left<u,alpha vright> = alpha left<u,vright>, quad left<alpha u,vright> = bar alpha left<u,vright>,
$$
where $bar alpha$ is the complex conjugate of $alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).
answered Sep 10 at 19:27
Kusma
3,440219
As you said, the notation $acdot b$ suggests bilinearity. The notations $left<a,bright>$ or $left<amid bright>$ or $left(a,bright)$ are better for non-bilinear scalar products, as those typically used in complex vector spaces, where you are as least as interested in sesquilinear forms that are linear in one component and antilinear in the other. For example,
$$left<u,alpha vright> = alpha left<u,vright>, quad left<alpha u,vright> = bar alpha left<u,vright>,
$$
where $bar alpha$ is the complex conjugate of $alpha$. (Whether the scalar product should be linear in the left or in the right component is a question of taste and of whether you are a physicist).
answered Sep 10 at 19:27
Kusma
3,440219
answered Sep 10 at 19:27
Kusma
3,440219
answered Sep 10 at 19:27
Kusma
3,440219
answered Sep 10 at 19:27
Kusma
3,440219
3,440219
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
add a comment |Â
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
The notations $acdot b$ and $langle a,brangle$ both look symmetric. Perhaps a better notation for $mathbb C$ spaces would be $a,lrcorner,b$ or $(a,brangle$ or something. Also, there is a philosophical question of whether complex numbers "should be" scalars; $mathbb R$eal timelike vectors act like imaginary units.
– mr_e_man
Sep 10 at 19:52
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
One interesting solution is to use two-sided scalar multiplication: $cv=vc^*$. Now we can have it both ways: $$c(vcdot w)=(cv)cdot w=vcdot(c^*,w)$$ $$(vcdot w)c=vcdot(wc)=(vc^*)cdot w$$ This is natural when described with a Clifford algebra (over Lorentzian $mathbb R^n,1$) model of the complex inner product space. (There may be other real models.)
– mr_e_man
Sep 11 at 14:34
add a comment |Â
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The notation $langle a | b rangle$ is quite practical in QM, since it allows you to really emphasize the duality between bras $langle a | $ and kets $| b rangle$. Other than that, whenever I see $a cdot b$ I think of the Euclidean scalar product. I generally use $langle a, b rangle$ when working with general inner products.
– Sobi
Sep 10 at 18:38
On the other hand $acdot b$ can denote so many different algebra products, e.g., see here. Of course we can use $ab$, $acirc b$, or $abullet b$ etc. You see, the problem can also be the other way around.
– Dietrich Burde
Sep 10 at 18:42
So many different names? Surely because it arises in so many different contexts.
– Lord Shark the Unknown
Sep 10 at 18:43
@LordSharktheUnknown -- Addition arises in even more contexts, but we always use $+$.
– mr_e_man
Sep 10 at 19:14
$a^Tb$ is the expression of a particular inner product relative to an specific, but often tacitly-understood, basis.
– amd
Sep 10 at 20:20