Does anyone know what C^3 take away the hyper planes defined by x=y, x=z and y=z...

Does anyone know what C^3 take away the hyper planes defined by x=y, x=z and y=z. I honestly have no clue how to even go about this.

By C^3 do you mean the complex vector space?

>x=y, x=z and y=z

Do you mean 3 subspaces for R 3?

And what the fuck do you mean by 'take away the'. Do you mean the '-' operation in set theory?

Then I may be able to answer.

You know what, just fuck it. Can you post the question in mathematical notation instead of bumfuck stoned descriptions?

I dont know how to use the notation in Veeky Forums but for your questions.

Yes the complex plane.

No I mean the sub spaces defined by the equations. For x=y its the subspace defined by vectors of the form (x,x,z) where x and z are complex numbers.

Lastly yes I ment the set theoretical '-'.

Sorry for the confusion.

Okay, for my own purposes I will be working over the reals. C^3 is isomorphic to R^6, whatever.

The equations x=y determine the subspace (a.b,a,b,c,d) where all those variables are real numbers.

x=z determines (a,b,c,d,a,b)

y=z determines (c,d,a,b,a,b)

Note that these subspaces are all isomorphic to R^4

If you join them all together you get the subspace (a,b.c,d,e,f) where either (a,b) = (c,d) or (c,d) = (e,f) or (a,b) = (e,f)

If any of those conditions is true then that vector is where we want it.

I don't even know if I'm right but my intuition tells me this new, bigger, subspace is also isomorphic to R^4

Now, because C^3 is our universe then the '-' operation relative to it is the same as getting the complement of the set we are removing, in this universe.

So we want the elements of C^3 that are NOT in the subspace I just described, so all we need to do is negate the condition of the subspace. Some propositional logic and one tautology later we arrive at the subspace

(a.b.c,d,e,f) where (a,b) != (c,d) AND (c,d) != (e,f) AND (a,b) != (e,f)

where != means not equal, in case you don't know C++.

Now, I don't even know how to imagine this subspace and I don't even know what is it isomorphic to so that I can make a comparison, but I do know that (a,b,c,d,e,f) is only an element of it if and only if (a,b) != (c,d) AND (c,d) != (e,f) AND (a,b) != (e,f)

Maybe because we are just taking 4 dimensional 'cross-sections' out of a 6 dimensional space this is isomorphic to C^3 but I have no idea.

By the way, I don't even know if I'm right. My memory of linear algebra is blurry.

Everything you said was true, also its good that you compared it to R^6 since its easy to think that taking a hyper plane from C^3 makes it disconnected but seeing it from the point of view of R^6 you can see that it is connected.


Also another similar example would be C^3 take away x=y y=z. This is isomorphic to C^* \times C^* \times C where C^* is C-{0}. Just for what I'm doing, I need to be able to extend this to a this system of 3 equalities (I dont know a better way of putting it).

You just described it. What kind of answer are you looking for?

>Does anyone know what the set {2,5,8} is?

I was hoping for a familiar/ nicer space. Kind of like the example in

>this new, bigger, subspace is also isomorphic to R^4

It's not. In fact, it's not even a subspace.
For example, (1,2, 0,0,1,2) and (1,1,1,1,0,0) are both in the subset, but their sum (2,3,1,1,1,2) is not.

>C^* \times C^* \times C where C^* is C-{0}

surround these things with [*math] [*/math] without the * like

[math]C^* \times C^* \times C[/math]

Anyways, thinking back to what a wrote, the set from my final result does not contain the zero vector so it isn't a subspace, just a subset.

And it is not an easy fix as I also noticed that it is not closed under addition, but I suppose thinking in terms of vector spaces is not necessary, I just fucked up by calling it a subspace.

>C^3 is isomorphic to R^6

Only if you disregard the fucking underlying algebraic structures.

>tfw normies learn about bijections and start calling them isomorphisms
Kill me now.

Yeah, I was too quick to give these subsets cute names. Hopefully my final result is still current as long as you call it a subset.

But we do not need to think of it as a group or anything of that nature for what OP is asking.

He wants to know what is the geometric result of taking those specific elements out of the entire set.

By noting that C^3 is isomorphic to R^6 as vector spaces I could easily visualize it and then start cutting it down to see what remained. That is all I need to know, what elements remain. Nothing to do with algebra here.

ayy fuck you I already posted about my mistake. I got carried away in terms of vector spaces and kept ignoring the details.

>In case you don't know C++
What about someone who knows one of the MANY languages that uses this notation but not C++? Do you think they know what you mean??? D:
I guess you ONLY know C++ since you thought that the notation was exclusive to C++.
I'm sorry I keep replying to you but this is really rustling my jimmies.

Also, subspace =|= (not equal just in case you can't read) subset.

>this is about vector spaces, not algebra!
A vector space is an algebraic structure, you retard.

>Das not what OP ask!
Well thats true about any question since OP never posed a question.

I know that C, C# and Java uses it. Probably any other C derived language.

I just said C++ because it is the only programming language that actually matters and I'd like to live in a world where only it existed.

Bijections are isomorphisms in Set.

There is no problem with this statement. They are both real vector spaces, and are isomorphic in that category. They are also both complex vector spaces and are isomorphic in that category as well.

>A vector space is an algebraic structure, you retard.

Yeah... and I already admitted my mistake about thinking of this problem in terms of vector spaces.

I don't even know what the fuck happened, literally just elementary set theory over here.

>Well thats true about any question since OP never posed a question.

Then what did I answer?

That's actually more than they said. A better example would be
>does anyone know what the set {1,3,2}?

kek'd

originalio commentario

R^6 n C^3 is more than simple set

>By noting that C^3 is isomorphic to R^6 as vector spaces I could easily visualize it and then start cutting it down to see what remained. That is all I need to know, what elements remain. Nothing to do with algebra here.
>talks about algebraic methods in the first sentence
>nothing to do with algebra

OP did not ask any question at all btw. Idk why you assume he's asking about geometry.

>2016
>accepting the Axiom of Choice

>Well thats true about any question since OP never posed a question.
wow I am a retard. I ment to ask what this space is homeomorphic to. I don't know how i pressed submit without having that down.

>>nothing to do with algebra

I mean that there was no point in treating anything over there as more than a subset. It was so pointless that it even led me to call mere subsets, subspaces.

All that needed to be noted was the isomorphism between the two spaces so that I could work in big 6-tall vectors.

But as rings or fields, they are not. And that's why we don't just say "isomorphic"

Tbh I suck at topology so I can't help you. :(
For my sake, can you define the topology on each set?

That's going to get you into trouble if you pursue math. "Just a small difference" is simply "wrong" most of the time.

point taken, but

>fields

Sorry :/

Its just the subspace topology gotten from the standard topology.

Ty based user.

>R^6 is a complex vector space
elaborate pls

Its homeomorphic to R^6 and even though I was dumb and didnt include that in the OP it was true in the context I needed it in.

are you agreeing its not a complex vector space? because i don't see how it could be

It is but the complex vector space C^3 is homeomorphic to R^6. And I am asking a topological question, not an algebraic so it doesnt matter that C^3 is not isomorphic to R^6.

If you perform a change of basis, the hyperplanes you mention are just the (complex) coordinate axes x=0, y=0, z=0, which you can think of as the complex analog of the coordinates axes in R^3.
The union of these three hyperplanes is the space of solutions to the equation xyz=0, which is a hyperplane singularity with coordinate ring [math]\frac{\mathBB{C}[x,y,z]}{(xyz)}[/math].

They arnt the axis though you can see that the line x=y=z is in the intersection of the 3 hyperplanes, and the x,y,z axis only have a point in common. Also you cant think of this as R^3 as I previously said that R^3- a hyperplane is disconnected, while C^3 - a hyperplane is still connected.

It's late and I'm slipping up, I meant the orthogonal complements of the axes, which in R^3 are 3 normal planes that intersect in three lines.
Complexifying a variety has some topological implications, but you're mostly just doubling the dimension of all your objects (line becomes a plane, plane becomes 4-plane etc).

Funniest letter ever recived

You said it has topological implications and sadly I am asking if they are homeomorphic I was just too dumb to put it in the op.

It's fake retard