Math Challenge - May 2021

  • Challenge
  • Thread starter fresh_42
  • Start date
  • #51
218
27
Sure, but what if ##z\neq 0##?
then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
as we had,
$$\begin{align}

2x=z(a+b)\nonumber\\

2y=z(a-b)\nonumber

\end{align}$$
and putting ##2cz=(a^2+b^2)z^2## ⇒ ##z=\frac {2c} {a^2+b^2}##in these two equations, we get the above values of (x,y,z)
 
  • #52
218
27
Problem #12
$$y^2=x\cdot (x+1)\cdot (x+7)\cdot (x+8)$$
substitute ##x+4 \rightarrow t##
then the equation becomes,
$$\begin{align}
y^2&=(t-4)\cdot(t-3)\cdot(t+3)\cdot(t+4)\nonumber\\
y^2&=(t^2-16)\cdot(t^2-9)\nonumber
\end{align}$$
so, for the R.H.S to be a perfect square,
the only possibilities are ##t=3,4,5##

as for product of two numbers (say ##a,b##) to be a perfect square, the only possibilities are,
if ##a=b,a=0,b=0\space or\space a=l^2,b=m^2## (where l,m are any real numbers)

if we use ##t^2-16=t^2-9##, then we won't get any solutions, so we'll have to use
##t^2-16=l^2\space \text{&} \space t^2-9=m^2##
from this we get,
##t^2=16+l^2=m^2+9##
clearly ##t=5## is the only possibility as 3,4,5 are pythagorean triplets.

Also we can have either ##t^2-9=0## or ##t^2-16=0##
from this we get [edit] ##t=3,4,-3,-4## [edit]

So putting the obtained values back in ##t=x+4## we get [edit] ##x=-8,-7,-1,0,1## [edit]

So ordered pairs ##(x,y)## are [edit] ##(-8,0);(-7,0);(-1,0);(0,0);(1,144);(1,-144)## [edit]

*Edited the answer to include all values of (x,y)
 
Last edited:
  • #53
15,542
13,639
then,
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
as we had,
$$\begin{align}

2x=z(a+b)\nonumber\\

2y=z(a-b)\nonumber

\end{align}$$
and putting ##2cz=(a^2+b^2)z^2## ⇒ ##z=\frac {2c} {a^2+b^2}##in these two equations, we get the above values of (x,y,z)
This works only for ##a^2+b^2\neq 0##. Your first post was already correct, except for one special case.
 
  • #54
218
27
This works only for ab≠0.
Why? What is the problem if ab=0, we should still have
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

Edit: I see that if both a & b are 0 then this is wrong. So yes both a & b shouldn't be zero , but one of them can be right?
 
  • #55
218
27
Your first post was already correct, except one special case.
Is a=b=0 the special case you were talking about here?
 
  • #56
15,542
13,639
Is a=b=0 the special case you were talking about here?
Yes. ##a=b=c=0## and ##x=y=0## is a possibility, where ##z## doesn't have to be zero.
 
  • #57
218
27
Yes. ##a=b=c=0## and ##x=y=0## is a possibility, where ##z## doesn't have to be zero.
But if c=0 then z is also 0
 
  • #58
15,542
13,639
But if c=0 then z is also 0
No. If ##a=b=c=0## and ##x=y=0## then ##z=1## is a solution, as is any arbitrary value for ##z##.
 
  • #59
218
27
No. If ##a=b=c=0## and ##x=y=0## then ##z=1## is a solution, as is any arbitrary value for ##z##.
Yes I missed that, I was looking at this expression
$$(x,y,z) = \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$
Didn't even notice the question, my bad.
 
  • #60
334
44
Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?
 
  • #61
15,542
13,639
Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?
What is the difference?
 
  • Informative
Likes fishturtle1
  • #62
334
44
What is the difference?
I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
 
  • #63
15,542
13,639
I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
Yes. As functions, it is the composition, which in coordinates is matrix multiplication.
 
  • Informative
Likes fishturtle1
  • #64
334
44
One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))## the set of ##\mathbb{K}G## homomorphisms from ##V## to ##W##?
 
  • #65
15,542
13,639
One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, w))## the set of ##\mathbb{K}G## homomorphisms from ##V## to ##W##?
I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## means the homomorphisms of representations as defined. We have ##\rho: G \longrightarrow \operatorname{GL}(V)## and ##\tau : G \longrightarrow \operatorname{GL}(W)##. I do not see any functions from ##G## to ##\mathbb{K} .## The condition of the characteristic simply allows us to divide by ##|G|## in the symmetry operator.
 
  • Informative
Likes fishturtle1
  • #66
334
44
I don't see why you need the group field, because linearity of homomorphism spaces is all that is used. ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## means the homomorphisms of representations as defined. We have ##\rho: G \longrightarrow \operatorname{GL}(V)## and ##\tau : G \longrightarrow \operatorname{GL}(W)##. I do not see any functions from ##G## to ##\mathbb{K} .## The condition of the characteristic simply allows us to divide by ##|G|## in the symmetry operator.
that clears things up, thank you!
 
  • #67
334
44
\textbf{Claim 1.} ##\text{Sym}(\varphi)## is a linear map from ##V \to W##.

Proof:
First, we have ##\text{char} \mathbb{K} \not\vert \vert G \vert##. So, ##\vert G \vert \neq 0## and ##\frac{1}{\vert G \vert}## is defined. For each ##g \in G##, we have ##(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W##. Since ##W## is a vector space, it is closed under addition and scalar multiplication. So, ##(\text{Sym}\varphi)(v) \in W##. Next, we check that ##\text{Sym}(\varphi)## is ##\mathbb{K}##-linear. Let ##u, v \in V## and ##\lambda \in \mathbb{K}##. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows ##\text{Sym}\varphi## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 2.} The map ##\text{Sym}## is a ##\mathbb{K}##-linear mapping.

Proof:
Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)## and ##\lambda \in \mathbb{K}##. For any ##g \in G##, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}


Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude ##\text{Sym}## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 3.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace##

Proof:
##(\subseteq)##: Let ##\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. By definition, we have ##\theta \circ \rho(g) = \tau(g) \circ \theta## for all ##g \in G##. In particular, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta##.
\\

##(\supseteq)##: Suppose for all ##g \in G##, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta##. Multiplying both sides by ##\rho(g)##, we have ##\tau(g) \circ \theta = \theta \circ \rho(g)##. This shows ##\supseteq##.
[]


\textbf{Claim 4.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

Proof:
Consider the map that sends every element in ##V## to ##0##. This map is contained in ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. So, ##\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)##. Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}## and ##g \in G##.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, ##\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, ##\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. We can conclude ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

[]

\textbf{Claim 5.} ##\text{Sym}## is a projection onto ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##.

Proof:
Let ##\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Then ##\text{Sym}(\varphi) = \varphi##. In particular, ##\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)##. It follows that ##\text{Sym}^2 = \text{Sym}## on ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. This proves Claim 5.
[]
 
  • #68
15,542
13,639
problem 4
\textbf{Claim 1.} ##\text{Sym}(\varphi)## is a linear map from ##V \to W##.
(1) ##\operatorname{Im(Sym)} \subseteq \operatorname{Hom}(V,W).## You haven't pointed out that well-definition is one of the 5 claims, as has been already mentioned by @Infrared. But since you mentioned it implicitly in your proof, I take the following for well-definition, too. (2)
Proof:
First, we have ##\text{char} \mathbb{K} \not\vert \vert G \vert##. So, ##\vert G \vert \neq 0## and ##\frac{1}{\vert G \vert}## is defined. For each ##g \in G##, we have ##(\tau(g) \circ \varphi \circ \rho(g^{-1})(v) \in W##. Since ##W## is a vector space, it is closed under addition and scalar multiplication. So, ##(\text{Sym}\varphi)(v) \in W##. Next, we check that ##\text{Sym}(\varphi)## is ##\mathbb{K}##-linear. Let ##u, v \in V## and ##\lambda \in \mathbb{K}##. We have

\begin{align*}
(\text{Sym}\varphi)(u + v) & = \left(\frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1})\right) (u + v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ \rho(g^{-1}) (u + v)\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ \varphi \circ (\rho(g^{-1})(u) + \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \left(\tau(g) \circ (\varphi \circ \rho(g^{-1})(u) + \varphi \circ \rho(g^{-1})(v))\right) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})(u)) + (\tau(g) \circ + \varphi \circ \rho(g^{-1})(v)) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ \varphi \circ \rho(g^{-1})(u) + \frac{1}{\vert G \vert} \sum_{g \in G}\tau(g) \circ + \varphi \circ \rho(g^{-1})(v) \\
&= (\text{Sym}\varphi)(u) + (\text{Sym}\varphi)(v)\\
\end{align*}

Also,

\begin{align*}
(\text{Sym}\varphi)(\lambda v) &= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(\lambda v) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \lambda \left(\tau(g) \circ \varphi \circ \rho(g^{-1})(v)\right) \\
&= \lambda \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))(v)\right) \\
&= \lambda (\text{Sym}\varphi)(v) \\
\end{align*}

This shows ##\text{Sym}\varphi## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 2.} The map ##\text{Sym}## is a ##\mathbb{K}##-linear mapping.
Linearity. (3) This is basically clear from the definition.
Proof:
Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}(V, W)## and ##\lambda \in \mathbb{K}##. For any ##g \in G##, we have

\begin{align*}
\text{Sym}(\varphi + \sigma) &= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ (\varphi + \sigma) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G}\tau(g) \circ ((\varphi \circ \rho(g^{-1}) + (\sigma \circ \rho(g^{-1}))) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&=\frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) + \frac{1}{\vert G \vert}\sum_{g \in G} (\tau(g) \circ \sigma \circ \rho(g^{-1})) \\
&= \text{Sym}(\varphi) + \text{Sym}(\sigma)
\end{align*}

Also,

\begin{align*}
\text{Sym}(\lambda\varphi) &= \frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ (\lambda\varphi) \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert}\sum_{g \in G} k(\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= k\frac{1}{\vert G \vert}\sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= k \text{Sym}(\varphi) \\
\end{align*}

We may conclude ##\text{Sym}## is ##\mathbb{K}##-linear.
[]

\textbf{Claim 3.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) = \lbrace \theta : V \longrightarrow W \vert \forall_{g \in G} \tau(g) \circ \theta \circ \rho(g^{-1}) = \theta\rbrace##
This is no claim. It is the definition of a homomorphism of representations. You have to prove that it holds for the symmetry operator as we defined it, i.e. that all ##\operatorname{Sym}(\varphi )## "commute" with the representations. It is actually one of two points where an argument is necessary. (The projection is the other one.)
Proof:
##(\subseteq)##: Let ##\theta \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. By definition, we have ##\theta \circ \rho(g) = \tau(g) \circ \theta## for all ##g \in G##. In particular, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta \circ \rho(g) \circ \rho(g^{-1}) = \theta##.
\\

##(\supseteq)##: Suppose for all ##g \in G##, ##\tau(g) \circ \theta \circ \rho(g^{-1}) = \theta##. Multiplying both sides by ##\rho(g)##, we have ##\tau(g) \circ \theta = \theta \circ \rho(g)##. This shows ##\supseteq##.
[]


\textbf{Claim 4.} ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.
I actually did not count this as a claim, since the linear spaces are already included by definition. We have some homomorphisms with an additional condition
$$
\{\vartheta :V\longrightarrow W\,|\,\forall_{g\in G}\, : \,\tau(g)\circ\vartheta\circ \rho(g^{-1})=\vartheta \}
$$
and all homomorphisms ##\operatorname{Hom}(V,W)## on the other hand. That it is a subspace follows from the linearity in the condition, which is already contained in your claim (2) if you drop the sums.
Proof:
Consider the map that sends every element in ##V## to ##0##. This map is contained in ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. So, ##\emptyset \neq \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)) \subset \text{Hom}_{\mathbb{K}}(V,W)##. Let ##\varphi, \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W)), \lambda \in \mathbb{K}## and ##g \in G##.
\\

We have $$(\varphi + \sigma)\circ \rho(g) = (\varphi \circ \rho(g)) + (\sigma\circ \rho(g)) = (\tau(g) \circ \varphi) + (\tau(g) \circ \sigma) = \tau(g)(\varphi + \sigma)$$
So, ##\varphi + \sigma \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Also,
$$((\lambda \varphi) \circ \rho(g)) = \lambda(\varphi \circ \rho(g)) = \lambda (\tau(g) \circ \varphi) = (\lambda\tau(g)) \circ \varphi = \tau(g) \circ (\lambda \varphi)$$

So, ##\lambda \varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. We can conclude ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))## is a subspace of ##\text{Hom}_{\mathbb{K}}(V, W)##.

[]

\textbf{Claim 5.} ##\text{Sym}## is a projection onto ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##.

Proof:
Let ##\varphi \in \text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. Then ##\text{Sym}(\varphi) = \varphi##. In particular, ##\text{Sym}(\text{Sym}(\varphi)) = \text{Sym}(\varphi)##. It follows that ##\text{Sym}^2 = \text{Sym}## on ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##. This proves Claim 5.
[]
This is wrong. ##\operatorname{Sym}(\varphi ) \stackrel{i.g.}{\neq } \varphi .## You have to calculate that ##\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )##.

You missed both crucial points which actually require some "proof":
  • ##\tau(h)\circ\operatorname{Sym}(\varphi)=\operatorname{Sym}(\varphi)\circ \rho(h) ##
  • ##\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )##
 
  • Informative
Likes fishturtle1
  • #69
15,542
13,639
Correction to the last point: I overlooked that you have chosen ##\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))##. So ##\operatorname{Sym}(\varphi )=\varphi ## indeed, but why?
 
  • #70
334
44
Correction to the last point: I overlooked that you have chosen ##\varphi \in \operatorname{Hom}_\mathbb{K}((\rho,V),(\tau,W))##. So ##\operatorname{Sym}(\varphi )=\varphi ## indeed, but why?
I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.
 
  • #71
15,542
13,639
I think the calculation is

\begin{align*}
\text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\
&= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\
&= \varphi \\
\end{align*}

Edit: Also, thank you for the feedback and corrections.
Now for the last one: why does ##\operatorname{Sym}(\varphi )## commute with representation matrices?
 
  • Like
Likes fishturtle1
  • #72
334
44
For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)##

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}
 
  • #73
15,542
13,639
For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)##

Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ (\tau(g) \circ \varphi \circ \rho(g^{-1})) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \varphi \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1})) \circ \rho(h) \\
&= \left(\frac{1}{\vert G \vert} \sum_{g \in G} (\tau(g) \circ \varphi \circ \rho(g^{-1}))\right) \circ \rho(h) \\
&= \text{Sym}\varphi \circ \rho(h) \\
\end{align*}
Yes, but ##\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}(V,W)## and we want to show that actually ##\operatorname{Sym}\, : \,\operatorname{Hom}(V,W)\longrightarrow \operatorname{Hom}((\rho,V),(\tau,W))##. That is, we have an arbitrary homomorphism ##\varphi \, : \,V\longrightarrow W##, and only its image satisfies the additional condition, which we want to show.

##\operatorname{Sym}## is a projection, i.e. maps something from bigger to smaller.

Hint: We haven't used that ##\rho## and ##\tau## are representations, yet.
 
  • Informative
Likes fishturtle1
  • #74
334
44
Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that ##\rho## and ##\tau## are homomorphisms.
 
  • #75
15,542
13,639
Proof: We have

\begin{align*}
\tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(h) \circ \tau(h^{-1}g) \varphi \circ \rho(g^{-1}h) \\
&= \frac{1}{\vert G \vert} \sum_{h^{-1}g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \circ \rho(h) \\
&= \text{Sym}(\varphi) \circ \rho(h) \\
\end{align*}

In the above calculations, we used that ##\rho## and ##\tau## are homomorphisms.
... and the fact that left multiplication in a group is a bijection ##L_{h^{-1}}\, : \,g \longmapsto h^{-1}g##.

Just saying, because the difficulty of the problem was mainly to identify all those seemingly clear facts. It is sometimes more difficult to see what has to be shown than it is to show it.
 
  • Informative
Likes fishturtle1

Related Threads on Math Challenge - May 2021

Replies
93
Views
2K
Replies
56
Views
3K
Replies
86
Views
6K
Replies
102
Views
3K
Replies
67
Views
4K
  • Sticky
  • Last Post
2
Replies
35
Views
725
Replies
100
Views
2K
Replies
48
Views
6K
  • Last Post
3
Replies
61
Views
7K
Replies
61
Views
3K
Top