<The previous article in this series | The table of contents of this series |
, while real numbers are as imaginary as imaginary numbers
Topics
About:
high school mathematics
The table of contents of this article
Starting Context
Target Context
-
The reader will know how the complex numbers system models the reality.
Orientation
There is an article on becoming a benefactor of humanity by being a conduit of truths
Main Body
1: Are Imaginary Numbers Imaginary?
Special-Student-7-Hypothesizer
Any imaginary number is \(b i\) where \(b\) is any real number and \(i\) is the number such that \(i^2 = - 1\).
There are some (maybe most) people who believe that imaginary numbers are "imaginary".
Special-Student-7-Rebutter
That is a bad name ...
Special-Student-7-Hypothesizer
The objection is natural: "The name is suggesting that they are imaginary ...".
But the name is a remnant of a historical ignorance.
In fact, misnomers are not so rare even in mathematics; "congruent" is one of them.
Special-Student-7-Rebutter
What "ignorance"?
Special-Student-7-Hypothesizer
Probably, the concept of 'imaginary number' was 1st conceived in the attempt to solve cubic equations.
Special-Student-7-Rebutter
"cubic"? instead of quadratic?
Special-Student-7-Hypothesizer
Probably so: you may think that the attempt to solve \(x^2 = - 1\) caused \(i\), but \(x^2 = - 1\) was just "It has no solution" without any further development.
But Cardano's method for cubic equations required \(\omega := (- 1 + \sqrt{3} i) / 2\), which is a cubic root of \(1\): you will see that \(\omega\) is indeed a cubic root of \(1\) by diligently expanding \(\omega^3\) with \(i^2 = - 1\).
But \(i\) at that time was just an expedient: "There is not such a thing, but if we pretend that it existed, ta-da!, we can solve cubic equations.".
Special-Student-7-Rebutter
"ta-da"?
Special-Student-7-Hypothesizer
Anyway, it was thought imaginary, meaning that "it does not really exist.".
Special-Student-7-Rebutter
To be accurate, it does not exist in the real numbers set.
Special-Student-7-Hypothesizer
Certainly, but the people at that time could not imagine beyond the real numbers set, and so, as far as it does not exist in the real numbers set, "it did not exist at all" for them.
Special-Student-7-Rebutter
Just as "\(\sqrt{2}\) did not exist at all" for the people who could not imagine beyond the rational numbers set.
Special-Student-7-Hypothesizer
Nothing was wrong with \(i\), but something was wrong with the people's imagination.
Special-Student-7-Rebutter
In the 1st place, what do "real" and "imaginary" mean?
2: Does Mathematics Discover or Invent?
Special-Student-7-Hypothesizer
In order to think of that, we should think of the question, "Does mathematics discover or invent?".
Special-Student-7-Rebutter
Ah, that is a well-heard question.
Special-Student-7-Hypothesizer
While some people have expressed their own stances, I am pretty sure on that question: mathematics is a model of the reality.
Special-Student-7-Rebutter
So, ..., does it discover or invent?
Special-Student-7-Hypothesizer
Any model itself is a human-made artifact, but mathematics is not something arbitrarily trumped-up but something that aspires to appropriately represent the reality.
Special-Student-7-Rebutter
Cannot a branch of mathematics not aspire so?
Special-Student-7-Hypothesizer
It could, but any branch gets some significant attention only when it represents the reality well, because otherwise, the branch would have no application.
Special-Student-7-Rebutter
So, ..., does mathematics discover or invent?
Special-Student-7-Hypothesizer
Mathematics models the reality, I am saying.
In fact, I do not think so meaningful to categorize it in "discover" or "invent": it is in "discover" in order to model the reality and is also in "invent" because it is a human-artifact.
3: Then, What Do "Real" and "Imaginary" Mean?
Special-Student-7-Hypothesizer
If "real" means being an entity in the reality, any concept in mathematics is not real, because the concept is an entity in the human-made model, not in the reality.
So, 'real numbers' will be no more real than 'imaginary numbers'.
Special-Student-7-Rebutter
It is not that real numbers exist in the reality.
Special-Student-7-Hypothesizer
But the real numbers set models the reality well.
In fact, the real numbers set models any line well.
Special-Student-7-Rebutter
It is not only spacial lines that the real numbers set models.
Special-Student-7-Hypothesizer
Certainly, for example, also masses of objects are modeled by real numbers.
Anyway, the real numbers set is regarded to be real because it models the reality well.
So, "real" is appropriately understood as modeling the reality well.
So, whether 'imaginary number' is real is the matter of whether it models the reality well.
4: How Complex Numbers Set Models Reality
Special-Student-7-Hypothesizer
The reason why 'imaginary number' was regarded to be non-real is that the people thought that it did not represent anything in the reality.
In fact, what line segment has the i length?
Special-Student-7-Rebutter
But a concept does not need to model a line but to model an entity in the reality.
Special-Student-7-Hypothesizer
In fact, we should think of the complex numbers set, of which the imaginary numbers set is a part.
The complex numbers set models any 2-dimensional plane.
For any 2-dimensional plane, any point is represented by a complex number, \(x + y i\), and any complex number represents a point on the plane; technically, the points set on the plane corresponds to the complex numbers set bijectively.
Then, \(i\) represents the \(0 + 1 i\) point on the plane.
Regarding \(x + y i\) as the vector from the origin to \(x + y i\), \(x + y i = r (cos \theta + sin \theta) i\), where \(r\) is the length of the vector and \(\theta\) is the counter-clock-wise-angle of the vector from the x-axis, as is well known.
Let us define \(e^{\theta i} := cos \theta + sin \theta i\).
Special-Student-7-Rebutter
That definition is possible only because \(e^{0 i} := cos 0 + sin 0 i = 1 = e^0\).
Special-Student-7-Hypothesizer
Well, I could insist that the exponential function for real numbers and the exponential function for imaginary numbers were totally different creatures with the same symbol happened to be shared, but that would be a bad practice, so, yes, we use the same symbol only because the 2 exponential functions are seamlessly connected.
Anyway, that is just a definition, because the exponential map with imaginary number arguments had not been defined as far as we were concerned.
Special-Student-7-Rebutter
The essence of the complex numbers set is not only that the set bijectively corresponds to the plane points set, set-wise.
In fact, what does \(i^2 = - 1\) mean?
Special-Student-7-Hypothesizer
Thinking of \(i^2\) means thinking of multiplication of complex numbers.
As the real numbers set allows the arithmetic operations, the complex numbers set allows the arithmetic operations, \((x + y i) + (x' + y' i) = (x + x') + (y + y') i\); \((x + y i) - (x' + y' i) = (x - x') + (y - y') i\); \((x + y i) (x' + y' i) = (x x' - y y') + (x y' + y x') i\); \((x + y i) / (x' + y' i) = ((x + y i) (x' - y' i)) / (x'^2 + y'^2) = ((x x' + y y') + (- x y' + y x') i) / (x'^2 + y'^2) = (x x' + y y') / (x'^2 + y'^2) + (- x y' + y x') / (x'^2 + y'^2) i\), where \(x' + y' i \neq 0\).
Special-Student-7-Rebutter
An important fact is that that is an expansion of the arithmetic operations for the real numbers set.
Special-Student-7-Hypothesizer
Yes, we have defined the arithmetic operations for the complex numbers set as they are consistent with the arithmetic operations for the real numbers set with the real numbers set as the subset of the complex numbers set.
Special-Student-7-Rebutter
So, the real numbers set sits seamlessly in the complex numbers set.
Special-Student-7-Hypothesizer
And the amazing fact is that \(e^{(\theta + \theta') i} = e^{\theta i} e^{\theta' i}\) holds: \(e^{(\theta + \theta') i} := cos (\theta + \theta') + sin (\theta + \theta') i\) while \(e^{\theta i} e^{\theta' i} = (cos \theta + sin \theta i) (cos \theta' + sin \theta' i) = (cos \theta cos \theta' - sin \theta sin \theta') + (cos \theta sin \theta' + sin \theta cos \theta') i = cos (\theta + \theta') + sin (\theta + \theta') i\).
And the exponential function for complex numbers can be seamlessly defined as \(e^{x + y i} := e^x e^{y i}\), satisfying \(e^{(x + y i) + (x' + y' i)} = e^{x + y i} e^{x' + y' i}\).
Special-Student-7-Rebutter
The definition of the exponential function for complex numbers is just a definition, but the amazing fact is that the function has become seamless with the exponential function for real numbers and satisfies the expected property.
Special-Student-7-Hypothesizer
As each \(x + y i\) is regarded to be a vector, the addition of any 2 complex numbers is the addition of the corresponding 2 vectors, as is obvious.
The product of any 2 complex numbers is the vector whose length is the product of the lengths of the 2 vectors and whose angle is the sum of the angles of the 2 vectors, which is because \((x + y i) (x' + y' i) = r e^{\theta i} r' e^{\theta' i} = r r' e^{(\theta + \theta') i}\).
Special-Student-7-Rebutter
So, the complex numbers set models the plane in the reality well.
Special-Student-7-Hypothesizer
So, the complex numbers set is no less real than the real numbers set.
5: Thinking Imaginary Numbers Imaginary Is a Sure Sign of Narrow-Mindedness
Special-Student-7-Hypothesizer
So, \(i\) is the \((0, 1)\) point on the plane and \(i^2 = - 1\) is reasonably understood as the result of the multiplication: \(i^2 = i i\) has the length, \(1 \times 1 = 1\), and the angle, \(\pi / 2 + \pi / 2 = \pi\), which is nothing but \(- 1\).
So, \(i^2 = - 1\) is no mystery at all.
Special-Student-7-Rebutter
Someone who thinks that \(i\) is imaginary is a 1-dimensional creature that is caged in a line and cannot see or imagine the existence of the world outside the line.
Special-Student-7-Hypothesizer
The word, "imaginary number", was coined at the time when the people did not understand how imaginary numbers modeled the reality, but much time has passed, many things have developed, and you should not be saying like "Imaginary numbers are imaginary." any more.
I heard an argument that wave function in quantum mechanics was not real because it was 'complex numbers'-valued, but that is a sure sign of ignorance.
Special-Student-7-Rebutter
Of course, everyone (especially, we) is ignorant more or less, and it will be fine if you humbly admit your ignorance and continually rectify your misconceptions.
6: Why We Think of 'Field'
Special-Student-7-Hypothesizer
The complex numbers set with the arithmetic operations constitutes a field, as the real numbers set with the arithmetic operations constitutes a field.
Also the rational numbers set with the arithmetic operations is a field, but the integers set with the addition, the subtraction, and the multiplication is not any field but a ring, which is because it does not have any division, which means that \(1 / 2\) does not belong to the integers set.
Special-Student-7-Rebutter
As has been mentioned in What Is a Vector?, it is important to think of the space.
Special-Student-7-Hypothesizer
You may think like "The division operation exists in the integers set, because we can do '1 / 2'", but that is not on the integers set because the result is not in the integers set.
Special-Student-7-Rebutter
Anyway, why should we introduce such an abstract concept as 'field'?
Special-Student-7-Hypothesizer
A major reason is, that is for the sake of economy.
In fact, as far as a structure is a field, some many conclusions hold for the structure regardless of that it is the rational numbers field, the real numbers field, the complex numbers field, or any other field.
So, it is very uneconomical to state and prove the conclusions individually for each of the various fields.
There seem some people who show sheer rejections to abstract concepts, but do you really want to individually deal with each of the various fields?
If you begin to individually study various fields, probably (unless you are too unintelligent), you will begin to feel studying essentially same things again and again, and probably (unless you are too unintelligent), you will find it unwise.
On the other hand, once you accept the concept of 'field', you can just study 'field' and can make same conclusions for all the fields.
Which do you prefer?
7: Why Not Higher-Dimensional Numbers Systems?
Special-Student-7-Rebutter
The complex numbers system (field) is fine modeling the 2-dimensional plane, but then, a natural question will be "Are there a numbers system for the 3-dimensional space, a numbers system for the 4-dimensional space, etc.?".
Special-Student-7-Hypothesizer
That question is "While the complexes numbers system exists, why not a numbers system like \(x + y i + z j\)? or does it exist?".
The answer is that it does not exist as a field.
Special-Student-7-Rebutter
Why?
Special-Student-7-Hypothesizer
If you define it just as a vectors space, it is of course possible: let \(x + y i + z j\) represent the \((x, y, z)\) point.
But it is not possible to define any set of arithmetic operations on it to make it a field.
Special-Student-7-Rebutter
Why?
Special-Student-7-Hypothesizer
It is just impossible: supposing that it expands the complex numbers field, setting \(i j = a + b i + c j\) where \(a, b, c\) are some real numbers, \(i i j = i (a + b i + c j)\), but the left hand side is \(i^2 j = - j\) and the right hand side is \(a i + b i^2 + c i j = a i - b + c i j = a i - b + c (a + b i + c j) = a c - b + (a + b c) i + c^2 j\), which means that \(a c - b = 0\), \(a + b c = 0\), and \(c^2 = - 1\), which is impossible because \(c^2 = - 1\) is impossible.
Special-Student-7-Rebutter
What if it does not expand the complex numbers field?
Special-Student-7-Hypothesizer
Usually, it is useful only because it expands the complex numbers field, as the complex numbers field is useful only because the complex numbers field has expended the real numbers field.
Special-Student-7-Rebutter
Anyway, what if?
Special-Student-7-Hypothesizer
Anyway, it should be impossible, although I do not show the proof here.
Special-Student-7-Rebutter
Fascinating! Why are the 1-dimensional space and the 2-dimensional space so special to allow the field structures?
Special-Student-7-Hypothesizer
Probably, the concept of 'field' is so special to choose only the 1-dimensional space and the 2-dimensional space.
Special-Student-7-Rebutter
What if the structure is not required to be 'field'?
Special-Student-7-Hypothesizer
In fact, there is 'quaternion' for the 4-dimensional space, which is not any field but an associative division algebra over the real numbers field, where "division" means that each nonzero element has the multiplicative inverse.
Special-Student-7-Rebutter
How is that "associative division algebra" thing different from 'field'?
Special-Student-7-Hypothesizer
That thing is not commutative: \(i j = k\) but \(j i = - k\), for example.
Special-Student-7-Rebutter
But there is no 3-dimensional associative division algebra over the real numbers field?
Special-Student-7-Hypothesizer
Certainly.
Special-Student-7-Rebutter
Fascinating! The 3-dimensional space has been skipped over by the structure.
Special-Student-7-Hypothesizer
It is interesting that while an \(m\)-dimensional Euclidean space and an \(n\)-dimensional Euclidean space do not seem so different as far as vectors space structure is concerned, they can be so different as far as field structure or associative division algebra structure is concerned.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series |
description/proof of that for sequence of points on metric space, sequence is Cauchy iff for each \(\epsilon\), there is \(N\) s.t. distance between \((N + 1)\)-th point and each subsequent point is smaller than \(\epsilon\)
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any sequence of points on any metric space, the sequence is Cauchy if and only if for each \(\epsilon\), there is an \(N\) such that the distance between the \((N + 1)\)-th point and each subsequent point is smaller than \(\epsilon\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the metric spaces }\}\)
\(s\): \(: \mathbb{N} \setminus \{0\} \to T\)
//
Statements:
\(s \in \{\text{ the Cauchy sequences }\}\)
\(\iff\)
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists N \in \mathbb{N} \setminus \{0\} (\forall j \in \mathbb{N} \setminus \{0\} \text{ such that } N \lt j (dist (s (N + 1), s (j)) \lt \epsilon)))\)
//
2: Proof
Whole Strategy: Step 1: suppose that there is such an \(N\); Step 2: see that \(s\) satisfies the Cauchy condition; Step 3: suppose that \(s\) satisfies the Cauchy condition; Step 4: see that there is such an \(N\).
Step 1:
Let us suppose that there is such an \(N\).
Step 2:
Let \(\epsilon\) be any.
For \(\epsilon / 2\), there is an \(N\) such that for each \(j\) such that \(N \lt j\), \(dist (s (N + 1), s (j)) \lt \epsilon / 2\).
For each \(l, m\) such that \(N \lt l, m\), \(dist (s (l), s (m)) \le dist (s (l), s (N + 1)) + dist (s (N + 1), s (m)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\), which means that \(s\) satisfies the Cauchy condition.
Step 3:
Let us suppose that \(s\) satisfies the Cauchy condition.
Let \(\epsilon\) be any.
There is an \(N\) such that for each \(l, m\) such that \(N \lt l, m\), \(dist (s (l), s (m)) \lt \epsilon\).
Taking \(l = N + 1\), \(dist (s (N + 1), s (m)) \lt \epsilon\), for each \(m\) such that \(N \lt m\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Lie group, differential of inversion at identity is inversion of argument
Topics
About:
group
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any Lie group, the differential of the inversion at the identity is the inversion of the argument.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the Lie groups }\}\)
\(i\): \(: G \to G\), \(= \text{ the inversion }\)
//
Statements:
\(d i_1: T_1M \to T_1M, v \mapsto - v\)
//
2: Proof
Whole Strategy: Step 1: let \(m\) be the multiplication map and let \(\phi: T_{(1, 1)}(G \times G) \to T_1G \times T_1G\) be the 'vectors spaces - linear morphisms' isomorphism, and think of \(m \circ (id, i): G \to G \times G \to G, g \mapsto m (id (g), i (g))\) and see that \(d (m \circ (id, i))_1 = d m_{(1, 1)} \circ \phi^{-1} \circ \phi \circ d (id, i)_1 = 0\); Step 2: see that \(\phi \circ d (id, i)_1 (v) = (v, d i_1 (v))\); Step 3: apply the proposition that for any Lie group and the 'vectors spaces - linear morphisms' isomorphism from the double-product \(C^\infty\) manifold tangent space at the identity onto the direct sum of the constituent tangent spaces at the identities, the composition of the differential of the multiplication at the identity after the inverse of the isomorphism is the addition of the arguments.
Step 1:
Let \(m: G \times G \to G\) be the multiplication map.
There is the 'vectors spaces - linear morphisms' isomorphism, \(\phi: T_{(1, 1)}(G \times G) \to T_1G \oplus T_1G, v \mapsto (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) where \(\pi_j: G \times G \to G\) is the projection into the \(j\)-th component, by the proposition that for any finite-product \(C^\infty\) manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.
Let us think of \(m \circ (id, i): G \to G \times G \to G, g \mapsto m (id (g), i (g))\) where \(id\) is the identity map.
\(m \circ (id, i)\) is \(C^\infty\) as a composition of \(C^\infty\) maps, by the proposition that for any maps between any arbitrary subsets of any \(C^\infty\) manifolds with boundary \(C^k\) at corresponding points, where \(k\) includes \(\infty\), the composition is \(C^k\) at the point.
\(d (m \circ (id, i))_1 = d m_{(1, 1)} \circ d (id, i)_1 = d m_{(1, 1)} \circ \phi^{-1} \circ \phi \circ d (id, i)_1 = 0\), because \(m \circ (id, i)\) is constant to \(1\).
Step 2:
For any \(v \in T_1G\), \(v\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G, t \mapsto \gamma (t)\).
\(\phi \circ d (id, i)_1 (v) = (d \pi_{1, (1, 1)} (d (id, i)_1 (v)), d \pi_{2, (1, 1)} (d (id, i)_1 (v))) = (d (\pi_1 ((id (\gamma (t)), i (\gamma (t))))) / d t \vert_0, d (\pi_2 ((id (\gamma (t)), i (\gamma (t))))) / d t \vert_0) = (d (id (\gamma (t))) / d t \vert_0, d (i (\gamma (t))) / d t \vert_0) = (d \gamma (t) / d t \vert_0, d (i (\gamma (t))) / d t \vert_0) = (v, d i_1 (v))\).
Step 3:
But \(d m_{(1, 1)} \circ \phi^{-1} ((v, d i_1 (v))) = v + d i_1 (v)\), by the proposition that for any Lie group and the 'vectors spaces - linear morphisms' isomorphism from the double-product \(C^\infty\) manifold tangent space at the identity onto the direct sum of the constituent tangent spaces at the identities, the composition of the differential of the multiplication at the identity after the inverse of the isomorphism is the addition of the arguments.
So, \(v + d i_1 (v) = 0\), so, \(d i_1 (v) = - v\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Lie group and 'vectors spaces - linear morphisms' isomorphism from double-product \(C^\infty\) manifold tangent space onto direct sum of constituent tangent spaces, composition of differential of multiplication at identity after inverse of isomorphism is addition of arguments
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any Lie group and the 'vectors spaces - linear morphisms' isomorphism from the double-product \(C^\infty\) manifold tangent space at the identity onto the direct sum of the constituent tangent spaces at the identities, the composition of the differential of the multiplication at the identity after the inverse of the isomorphism is the addition of the arguments.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the Lie groups }\}\)
\(\phi\): \(: T_{(1, 1)}(G \times G) \to T_1G \oplus T_1G\), \(= \text{ the 'vectors spaces - linear morphisms' isomorphism }\)
\(m\): \(: G \times G \to G\), \(= \text{ the multiplication map }\)
\(d m_{(1, 1)}\): \(: T_{(1, 1)}(G \times G) \to T_1G\), \(= \text{ the differential }\)
//
Statements:
\(d m_{(1, 1)} \circ \phi^{-1}: T_1G \oplus T_1G \to T_1G, (v_1, v_2) \mapsto v_1 + v_2\)
//
2: Proof
Whole Strategy: Step 1: see that \(\phi (v) = (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) and \(d m_{(1, 1)} \circ \phi^{-1}\) is linear; Step 2: see that \(d m_{(1, 1)} \circ \phi^{-1} ((v^1, 0)) = v^1\) and \(d m_{(1, 1)} \circ \phi^{-1} ((0, v^2)) = v^2\); Step 3: conclude the proposition.
Step 1:
For each \(v \in T_{(1, 1)}(G \times G)\), \(\phi (v) = (d \pi_{1, (1, 1)} (v), d \pi_{2, (1, 1)} (v))\) where \(\pi_j: G \times G \to G\) is the projection into the \(j\)-th component, by the proposition that for any finite-product \(C^\infty\) manifold with boundary, at each point of the manifold with boundary, there is a 'vectors spaces - linear morphisms' isomorphism from the tangent vectors space at the point onto the direct sum of the tangent vectors spaces of the constituents at the corresponding points.
As \(\phi\) is a 'vectors spaces - linear morphisms' isomorphism, \(d m_{(1, 1)} \circ \phi^{-1}\) is linear.
Step 2:
Let us think of any \((v^1, 0) \in T_1G \oplus T_1G\).
\(v := \phi^{-1} ((v^1, 0))\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G \times G, t \mapsto (\gamma^1 (t), 1)\), where \(v^1 = d \gamma^1 (t) / d t \vert_0\).
\(d m_{(1, 1)} (v) = d (m \circ \gamma (t)) / d t \vert_0 = d (\gamma^1 (t) 1) / d t \vert_0 = d \gamma_1 (t) / d t \vert_0 = v^1\).
That means that \(d m_{(1, 1)} \circ \phi^{-1} ((v^1, 0)) = v^1\).
Let us think of any \((0, v^2) \in T_1G \oplus T_1G\).
\(v := \phi^{-1} ((0, v_2))\) can be realized by a \(C^\infty\) curve, \(\gamma: I \to G \times G, t \mapsto (1, \gamma^2 (t))\), where \(v^2 = d \gamma^2 (t) / d t \vert_0\).
\(d m_{(1, 1)} (v) = d (m \circ \gamma (t)) / d t \vert_0 = d (1 \gamma^2 (t)) / d t \vert_0 = d \gamma^2 (t) / d t \vert_0 = v^2\).
That means that \(d m_{(1, 1)} \circ \phi^{-1} ((0, v_2)) = v_2\).
Step 3:
As \(d m_{(1, 1)} \circ \phi^{-1}\) is linear, \(d m_{(1, 1)} \circ \phi^{-1} ((v_1, v_2)) = d m_{(1, 1)} \circ \phi^{-1} ((v_1, 0) + (0, v_2)) = d m_{(1, 1)} \circ \phi^{-1} ((v_1, 0)) + d m_{(1, 1)} \circ \phi^{-1} ((0, v_2)) = v_1 + v_2\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of Lie subgroup
Topics
About:
group
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Lie subgroup.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( G'\): \(\in \{\text{ the Lie groups }\}\)
\(*G\): \(\in \{\text{ the subgroups of } G'\}\) with a topology and an atlas, \(\in \{\text{ the Lie groups }\}\)
//
Conditions:
\(G \in \{\text{ the immersed submanifolds of } G'\}\)
//
References
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definition of Lie group
Topics
About:
group
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Lie group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(*G\): \(\in \{\text{ the groups }\} \cap \{\text{ the } C^\infty \text{ manifolds } \}\)
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Conditions:
\(m: G \times G \to G, (g_1, g_2) \mapsto g_1 g_2 \in \{\text{ the } C^\infty \text{ maps }\}\), called "multiplication map"
\(i: G \to G, g \mapsto g^{-1} \in \{\text{ the } C^\infty \text{ maps }\}\), called "inversion map"
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References
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definition of finite-product Hilbert space
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of finite-product Hilbert space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\( \{V_1, ..., V_n\}\): \(\subseteq \{\text{ the } F \text{ Hilbert spaces }\}\), with any inner products, \(\{\langle \bullet, \bullet \rangle_1, ..., \langle \bullet, \bullet \rangle_n\}\)
\(*V_1 \times ... \times V_n\): \(= \text{ the product vectors space with the product inner product }\) with the metric induced by the norm induced by the inner product, \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
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Conditions:
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2: Note
Let us see that \(V_1 \times ... \times V_n\) is indeed a complete metric space, so, indeed a Hilbert space.
Let \(s: \mathbb{N} \to V_1 \times ... \times V_n\) be any Cauchy sequence.
For each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(m, o \in \mathbb{N}\) such that \(N \lt m, o\), \(\langle s (o) - s (m), s (o) - s (m) \rangle \lt \epsilon^2\).
\(\langle s (o) - s (m), s (o) - s (m) \rangle = \langle (s (o)^1, ..., s (o)^n) - (s (m)^1, ..., s (m)^n), (s (o)^1, ..., s (o)^n) - (s (m)^1, ..., s (m)^n) \rangle = \langle (s (o)^1 - s (m)^1, ..., s (o)^n - s (m)^n), (s (o)^1 - s (m)^1, ..., s (o)^n - s (m)^n) \rangle = \langle s (o)^1 - s (m)^1, s (o)^1 - s (m)^1 \rangle_1 + ... + \langle s (o)^n - s (m)^n, s (o)^n - s (m)^n \rangle_n \lt \epsilon^2\), which means that for each \(j \in \{1, ..., n\}\), \(\langle s (o)^j - s (m)^j, s (o)^j - s (m)^j \rangle_j \lt \epsilon^2\).
That means that \(s^j: \mathbb{N} \to V_j\) is a Cauchy sequence.
As \(V_j\) is complete, \(s^j\) converges to a \(v^j \in V_j\).
Let \(v := (v^1, ..., v^n) \in V_1 \times ... \times V_n\).
\(s\) converges to \(v\), because for each \(\epsilon\), we can take an \(N\) such that for each \(m\) such that \(N \lt m\) and each \(j\), \(\langle v^j - s (m)^j \rangle \lt \epsilon^2 / n\), and then, \(\langle v - s (m), v - s (m) \rangle = \langle (v^1, ..., v^n) - (s (m)^1, ..., s (m)^n), (v^1, ..., v^n) - (s (m)^1, ..., s (m)^n) \rangle = \langle v^1 - s (m)^1, v^1 - s (m)^1 \rangle_1 + ... \langle v^n - s (m)^n, v^n - s (m)^n \rangle_n \lt \epsilon^2 / n + ... + \epsilon^2 / n = \epsilon^2\).
References
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description/proof of that for finite-product vectors space with finite-product inner product, topology induced by product inner product is product topology of topologies induced by inner products
The table of contents of this article
Main Body
References
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definition of finite-product inner product
Topics
About:
vectors space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of finite-product inner product.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( F\): \(\in \{ \mathbb{R}, \mathbb{C} \}\), with the canonical field structure
\( \{V_1, ..., V_n\}\): \(\subseteq \{\text{ the } F \text{ vectors spaces }\}\), with any inner products, \(\{\langle \bullet, \bullet \rangle_1, ..., \langle \bullet, \bullet \rangle_n\}\)
\( V_1 \times ... \times V_n\): \(= \text{ the product vectors space }\)
\(*\langle \bullet, \bullet \rangle\): \(: (V_1 \times ... \times V_n) \times (V_1 \times ... \times V_n) \to F\), \(\in \{\text{ the inner products for } V_1 \times ... \times V_n\}\)
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Conditions:
\(\forall (v_1, ..., v_n), (v'_1, ..., v'_n) \in V_1 \times ... \times V_n (\langle (v_1, ..., v_n), (v'_1, ..., v'_n) \rangle = \langle v_1, v'_1 \rangle_1 + ... + \langle v_n, v'_n \rangle_n)\)
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2: Note
Let us see that \(\langle \bullet, \bullet \rangle\) is indeed an inner product.
Let \(v_1 = (v_{1, 1}, ..., v_{1, n}), v_2 = (v_{2, 1}, ..., v_{2, n}), v_3 = (v_{3, 1}, ..., v_{3, n}) \in V_1 \times ... \times V_n\) be any; let \(r_1, r_2 \in F\) be any.
1) \((0 \le \langle v_1, v_1 \rangle)\) \(\land\) \((0 = \langle v_1, v_1 \rangle \iff v_1 = 0)\): \(0 \le \langle v_{1, 1}, v_{1, 1} \rangle_1 + ... + \langle v_{1, n}, v_{1, n} \rangle_n = \langle (v_{1, 1}, ..., v_{1, n}), (v_{1, 1}, ..., v_{1, n}) \rangle = \langle v_1, v_1 \rangle\); when \(0 = \langle v_1, v_1 \rangle\), each \(\langle v_{1, j}, v_{1, j} \rangle_j = 0\), so, each \(v_{1, j} = 0\), so, \(v_1 = (v_{1, 1}, ..., v_{1, n}) = 0\); when \(v_1 = (v_{1, 1}, ..., v_{1, n}) = 0\), each \(v_{1, j} = 0\), so, each \(\langle v_{1, j}, v_{1, j} \rangle_j = 0\), so, \(0 = \langle v_1, v_1 \rangle\).
2) \(\langle v_1, v_2 \rangle = \overline{\langle v_2, v_1 \rangle}\), where the over-line denotes the complex conjugate: \(\langle v_1, v_2 \rangle = \langle v_{1, 1}, v_{2, 1} \rangle_1 + ... + \langle v_{1, n}, v_{2, n} \rangle_n = \overline{\langle v_{2, 1}, v_{1, 1} \rangle_1} + ... + \overline{\langle v_{2, n}, v_{1, n} \rangle_n} = \overline{\langle v_{2, 1}, v_{1, 1} \rangle_1 + ... + \langle v_{2, n}, v_{1, n} \rangle_n} = \overline{\langle v_2, v_1 \rangle}\).
3) \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_2, v_3 \rangle\): \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = \langle r_1 (v_{1, 1}, ..., v_{1, n}) + r_2 (v_{2, 1}, ..., v_{2, n}), (v_{3, 1}, ..., v_{3, n}) \rangle = \langle (r_1 v_{1, 1} + r_2 v_{2, 1}, ..., r_1 v_{1, n} + r_2 v_{2, n}), (v_{3, 1}, ..., v_{3, n}) \rangle = \langle r_1 v_{1, 1} + r_2 v_{2, 1}, v_{3, 1} \rangle_1 + ... + \langle r_1 v_{1, n} + r_2 v_{2, n}, v_{3, n} \rangle_n = r_1 \langle v_{1, 1}, v_{3, 1} \rangle_1 + r_2 \langle v_{2, 1}, v_{3, 1} \rangle_1 + ... + r_1 \langle v_{1, n}, v_{3, n} \rangle_n + r_2 \langle v_{2, n}, v_{3, n} \rangle_n = r_1 (\langle v_{1, 1}, v_{3, 1} \rangle_1 + ... + \langle v_{1, n}, v_{3, n} \rangle_n) + r_2 (\langle v_{2, 1}, v_{3, 1} \rangle_1 + ... + \langle v_{2, n}, v_{3, n} \rangle_n) = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_2, v_3 \rangle\).
The product needs to be finite, because otherwise, the inner product may not be into \(F\).
References
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definition of group right action that corresponds to group left action
Topics
About:
group
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of group right action that corresponds to group left action.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( G\): \(\in \{ \text{ the groups } \}\)
\( S\): \(\in \{ \text{ the sets } \}\)
\( f\): \(: G \times S \to S\), \(\in \{\text{ the group left actions }\}\)
\(*f'\): \(: S \times G \to S, (s, g) \mapsto f (g^{-1}, s)\), \(\in \{\text{ the group right actions }\}\)
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Conditions:
//
2: Note
Let us see that \(f'\) is indeed a group right action.
For each \(g_1, g_2 \in G\) and each \(s \in S\), \(f' (f' (s, g_1), g_2) = f' (f (g_1^{-1}, s), g_2) = f (g_2^{-1}, f (g_1^{-1}, s)) = f (g_2^{-1} g_1^{-1}, s) = f ((g_1 g_2)^{-1}, s) = f' (s, g_1 g_2)\).
For each \(s \in S\), \(f' (s, 1) = f (1^{-1}, s) = f (1, s) = s\).
Note that \(f\) itself cannot be regarded to be any group right action; we have constructed the group right action from \(f\): 'group left action' is not about just denoting \(G \times S\) instead of \(S \times G\): \(f'': S \times G \to S, (s, g) \mapsto f (g, s)\) is not any group right action, because \(f'' (f'' (s, g_1), g_2) = f'' (f (g_1, s), g_2) = f (g_2, f (g_1, s)) = f (g_2 g_1, s) = f'' (s, g_2 g_1)\), which does not equal \(f'' (s, g_1 g_2)\) in general.
References
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