B Is imaginary "i" a purely aesthetic matter?

[Klystron]

On the Langauge sub-theme:

If one accepts that language derives from culture expressed in an environment -- Inuit language has n words for 'snow' comes to mind from Anthropology -- then the expression "Mathematics is the language of Science" not only describes math as a language but science as its environment (culture).

 

[WWGD]

On the Langauge sub-theme:

If one accepts that language derives from culture expressed in an environment -- Inuit language has n words for 'snow' comes to mind from Anthropology -- then the expression "Mathematics is the language of Science" not only describes math as a language but science as its environment (culture).

But can't the language aspect be studies separately and independently-from the content? We can make a purely syntactic analysis of Mathematical language as a first-order language. These discussions get weird after a while and seemingly drift of, so sorry if I am not making too much sense.

 

[PeroK]

Is it possible that are there other extensions to ℝ that provide solutions to other types of non-polynomial problems? For example, if I were to define a number j such that sin(j) = 2, everybody would most likely find that pretty useless (otherwise someone else would have thought of that already), but is it possible that there are other extensions beyond ℂ that would be useful for something, or do the mathematicians think that there's nothing beyond ℂ, and that's really the final frontier?

The importance of $\mathbb{C}$ is that it is an algebraically closed field. In terms of numbers, that's part of the reason that further extensions are not normally considered numbers. But, of course, you have complex vectors and complex matrices etc. And, as others have said, there are quaternions etc.

Note that there do exist solutions to $\sin z = 2$ for complex $z$.

Interestingly, from a physics point of view, Quantum Mechanics is built on complex vector spaces, and complex functions of a real variable. In general, although measurable quantities: position, momentum, spin etc. are real numbers, the underlying structure of QM is complex.

 

[fbs7]

Ohhhhhhh... so the keyword is "algebraically closed". Wow, I'm starting to get it! From wikipedia... "F is algebraically closed field if every non-constant polynomial has a root"... which it says it is equivalent to saying that "F has no proper extension".

That means that ℂ cannot be extended just by polynomials, correct? Does that mean that it can't be extended either with any function that can be expressed as infinite series? I mean, an infinite series is a polynomial, but it's infinite, so is it possible that ℂ can be extended through some non-polynomial function that cannot be expressed as a series?

This is fascinating stuff!

 

[wle]

One way/reason for defining i as it has been done is that it allows for finding roots of polynomial of the type x^2+b . Formally, this is a field extension R[X]/(x^2+1), from which it follows that x^2+1 =0 , so that x^2 = -1. This gels together when having x=i so that x^2 =-1. So the construction/definition of i is not, in this sense, just formal, aesthetic.

They really became important for expressing the solutions to cubic polynomials (see here). Basically, cubic polynomials always have (at least) one real root and it is possible to say how to compute it, but you sometimes need complex numbers in the intermediate steps. For example, cubic polynomials of the form $$x^{3} + p x + q$$ have a root given by the Cardano formula, $$x = \Biggl(- \frac{q}{2} + \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}\Biggr)^{1/3} + \Biggl(- \frac{q}{2} - \sqrt{\frac{q^{2}}{4} + \frac{p^{3}}{27}}\Biggr)^{1/3} \,.$$ That root is real, but depending on the values of $p$ and $q$ it's possible for the term $\frac{q^{2}}{4} + \frac{p^{3}}{27}$ under the square roots to be negative. In that case the Cardano formula can still be meaningful and gives the correct (real) result, but only if you accept it containing intermediate expressions $\Bigl(a \pm b \sqrt{-1}\Bigr)^{1/3}$ with imaginary parts that end up cancelling out.

Interesting! I'd never run into this before.

Just to test this, I tried it on examples of addition and multiplication, using 2 + 3i and 5 - 2i
Addition
$\begin{bmatrix} 2 & -3 \\ 3 & 2\end{bmatrix} + \begin{bmatrix}5 & 2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix}7 & -1 \\ 1 & 7 \end{bmatrix}$
The last matrix represents 7 + 1i, which is the sum of 2 + 3i and 5 - 2i
Multiplication
$\begin{bmatrix} 2 & -3 \\ 3 & 2\end{bmatrix} \begin{bmatrix}5 & 2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix}16 & -11 \\ 11 & 16 \end{bmatrix}$
Same result that you get by multiplying (2 + 3i) and (5 - 2i).

Or if you're familiar with the properties of Pauli matrices, $$\begin{eqnarray*}
\bigl( a \, \mathbb{I} + i b \, \sigma_{y} \bigr) \bigl( c \, \mathbb{I} + i d \, \sigma_{y} \bigr) &=& ac \, \mathbb{I} + i (ad + bc) \, \sigma_{y} - b d \, {\sigma_{y}}^{2} \\
&=& (ac - bd) \, \mathbb{I} + i (ad + bc) \, \sigma_{y} \,,
\end{eqnarray*}$$ since ${\sigma_{y}}^{2}$ is the identity. (It doesn't matter if you use $\sigma_{y}$ or $-\sigma_{y}$.)

 

[fresh_42]

"F is algebraically closed field if every non-constant polynomial has a root"... which it says it is equivalent to saying that "F has no proper extension".

This is not true. There are extensions of $\mathbb{C}$. The division algebras $\mathbb{H}$ and $\mathbb{O}$ have already be named. But you can also add a new transcendental number, say $T$, so $\mathbb{C}[T]$ which is the same as the polynomial ring over $\mathbb{C}$ first becomes an integral domain, and thus has a quotient field $\mathbb{C}(T)$, which is a strict field extension of the complex numbers. And you can repeat this process as often as you want. The correct wording is: "... has no proper algebraic field extension!" Both restrictions are necessary as the examples I mentioned show.

 

[fresh_42]

Or if you're familiar with the properties of Pauli matrices, $$\begin{eqnarray*}
\bigl( a \, \mathbb{I} + i b \, \sigma_{y} \bigr) \bigl( c \, \mathbb{I} + i d \, \sigma_{y} \bigr) &=& ac \, \mathbb{I} + i (ad + bc) \, \sigma_{y} - b d \, {\sigma_{y}}^{2} \\
&=& (ac - bd) \, \mathbb{I} + i (ad + bc) \, \sigma_{y} \,,
\end{eqnarray*}$$ since ${\sigma_{y}}^{2}$ is the identity. (It doesn't matter if you use $\sigma_{y}$ or $-\sigma_{y}$.)

This is way over the top. To cite Pauli matrices just to have a name for $i \sigma_y$ is very far fetched. It's like starting with a factor group of $SO(4)$ just to explain a complex number. Sorry, but this is ridiculous.

 

[fbs7]

This is not true. There are extensions of $\mathbb{C}$. The division algebras $\mathbb{H}$ and $\mathbb{O}$ have already be named. But you can also add a new transcendental number, say $T$, so $\mathbb{C}[T]$ which is the same as the polynomial ring over $\mathbb{C}$ first becomes an integral domain, and thus has a quotient field $\mathbb{C}(T)$, which is a strict field extension of the complex numbers. And you can repeat this process as often as you want. The correct wording is: "... has no proper algebraic field extension!" Both restrictions are necessary as the examples I mentioned show.

Oh... got it (kinda)... I always keep omitting words that end up being relevant! Thanks for the correction!

 

[fbs7]

This is way over the top. To cite Pauli matrices just to have a name for $i \sigma_y$ is very far fetched. It's like starting with a factor group of $SO(4)$ just to explain a complex number. Sorry, but this is ridiculous.

WOW... sudden realization! This is exciting! For the last 20 years I meandered around them mysterious quantum formulas (without understanding anything, but fascinated on how cute they look) people would talk about SO(2), SU(3), etc... and refer to Lie algebra, multiplicative groups, ... ah!... no way someone with under 30 trillion neurons can understand any of that!

Now I see - SO(2) is the same thing as a 2D rotation in ℝ², and SO(3) is a 3D rotation in ℝ³. Hooray! So SO(4) may be related to 4D rotations? Or something else.. the wording is really difficult to translate a human English. Also I guess SU(3) too must be related to some other transformation-thingie with some particular property-thingie-thingie.

So SO(2) is related (related-sy, however the proper Mathematish phrase for that) to z ∈ ℂ such that ||z|| = 1! I learned something!

 

[fresh_42]

WOW... sudden realization! This is exciting! For the last 20 years I meandered around them mysterious quantum formulas (without understanding anything, but fascinated on how cute they look) people would talk about SO(2), SU(3), etc... and refer to Lie algebra, multiplicative groups, ... ah!... no way someone with under 30 trillion neurons can understand any of that!

Now I see - SO(2) is the same thing as a 2D rotation in ℝ², and SO(3) is a 3D rotation in ℝ³. Hooray! So SO(4) may be related to 4D rotations? Or something else.. the wording is really difficult to translate a human English. Also I guess SU(3) too must be related to some other transformation-thingie with some particular property-thingie-thingie.

So SO(2) is related (related-sy, however the proper Mathematish phrase for that) to z ∈ ℂ such that ||z|| = 1! I learned something!

Here's a list of "what is what" resp. "what an be viewed as what": https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
The Pauli matrices just give a simple basis for regular $2\times 2$ matrices with trace zero. They are intended to deal with $SU(2)$, although dropping the factor $i$ which in QFT is noted along $\hbar$. Therefore one could write $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ as needed in post #10 as $i\cdot \sigma_y$, but as I said, this is a bit too far fetched.

 

[sysprog]

WOW... sudden realization! This is exciting! For the last 20 years I meandered around them mysterious quantum formulas (without understanding anything, but fascinated on how cute they look) people would talk about SO(2), SU(3), etc... and refer to Lie algebra, multiplicative groups, ... ah!... no way someone with under 30 trillion neurons can understand any of that!

Now I see - SO(2) is the same thing as a 2D rotation in ℝ², and SO(3) is a 3D rotation in ℝ³. Hooray! So SO(4) may be related to 4D rotations? Or something else.. the wording is really difficult to translate a human English. Also I guess SU(3) too must be related to some other transformation-thingie with some particular property-thingie-thingie.

So SO(2) is related (related-sy, however the proper Mathematish phrase for that) to z ∈ ℂ such that ||z|| = 1! I learned something!

The membership roster of the SU(3) non-Abelian homology (gauge) group was used to (as it turns out, apparently correctly), predict the existence of the top quark in QCD, which helped to bring about the marshaling of greater effort to find it -- ephemeral, because it doesn't hadronize like the others, but part of the group, so the QCD theorists assiduously sought it.

 

[TeethWhitener]

So SO(2) is related (related-sy, however the proper Mathematish phrase for that) to z ∈ ℂ such that ||z|| = 1! I learned something!

Not just related. $SO(2)$ is isomorphic to the circle group, the unit circle in the complex plane. The representation that @mfb pointed out occurs because the set of all skew-symmetric matrices under standard matrix multiplication forms a Lie algebra, $o(2)$, with the commutator as the Lie bracket. The exponential map takes elements from this Lie algebra to the Lie group $SO(2)$, just as the exponential map takes elements from the Lie algebra $i\mathbb{R}$ to the Lie group called the circle group (and denoted by Wikipedia as $\mathbb{T}$).

 

[stevendaryl]

This is way over the top. To cite Pauli matrices just to have a name for $i \sigma_y$ is very far fetched. It's like starting with a factor group of $SO(4)$ just to explain a complex number. Sorry, but this is ridiculous.

Well, Hestenes had a program of replacing all occurrences of "i" in physics by elements of Clifford algebra. I don't know how successful his program was.

 

[Klystron]

Not just related. $SO(2)$ is isomorphic to the circle group, the unit circle in the complex plane. The representation that @mfb pointed out occurs because the set of all skew-symmetric matrices under standard matrix multiplication forms a Lie algebra, $o(2)$, with the commutator as the Lie bracket. The exponential map takes elements from this Lie algebra to the Lie group $SO(2)$, just as the exponential map takes elements from the Lie algebra $i\mathbb{R}$ to the Lie group called the circle group (and denoted by Wikipedia as $\mathbb{T}$).

Let me also call attention to a series of Insight articles beginning with
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/

Excellent exercise for aging brains <grin>.

 

[fresh_42]

Well, Hestenes had a program of replacing all occurrences of "i" in physics by elements of Clifford algebra. I don't know how successful his program was.

... and $\mathbb{C}$ is a real (associative) superalgebra with $\mathbb{C}_0=\mathbb{R}$ and $\mathbb{C}_1=i \cdot \mathbb{R}$. Super!