Math Challenge - May 2021

[benorin]

To start with separate the integrals like this

$$I=\int_0^\infty\int_0^\infty e^{-\tfrac{\lambda ^3}{xy}}x^{-\tfrac{2}{3}}y^{-\tfrac{1}{3}}\, dx\, dy \cdot \underbrace{\left( \int_0^\infty e^{-x}\, dx\right) ^2}_{=4 \Gamma ^2 (2)=4}$$

both converge as they should. Then to work on the left hand integrals by transformations. Will edit with the rest later.

 

[fresh_42]

This happens, dear students, if you do not note the dependencies! It has to be "for all $\varepsilon$ there is an $N(\varepsilon )$" and not "for all $\varepsilon$ there is an $N$". Sloppiness has its price!

 

[benorin]

I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows

 

[fresh_42]

I was just taking a pee and realized my freshman mistake lmao I only do math once a month now and it shows

But if you find the idea, then it is really beautiful. @Fred Wright has been close, even with the series, but this isn't the nice solution. Forget series.

 

[Infrared]

I think Problem 2 can be done without any fiddling with commutators:

Consider the function $\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.$ By the universal property of free groups, this uniquely extends to a homomorphism $F_n\to\mathbb{Z}^n.$ Note that $x\in F_n$ lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator $w_i$ is zero).

Since $\mathbb{Z}^n$ is abelian, this homomorphism descends to the quotient $F_n/[F_n,F_n]\to\mathbb{Z}_n.$ We check that this map is injective by constructing a left inverse: since both groups are abelian and $\mathbb{Z}^n$ is free abelian, the function $\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]$ taking $e_i$ to the class of $w_i$ (uniquely) extends to a homomorphism $\mathbb{Z}^n\to F_n/[F_n,F_n]$ which is clearly a left inverse. (In fact, the map $F_n/[F_n,F_n]\to\mathbb{Z}^n$ is also clearly surjective so it is an isomorphism).

Now, by injectivity, $\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].$ On the other hand, we already know that the kernel of $F^n\to\mathbb{Z}^n$ is the set of elements in $F_n$ whose exponents satisfy the given condition.

(Here, $e_i$ is the vector whose $i$-th component is $1$ and all others are zero.)

 

[Infrared]

Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that $\text{Sym}$ is well-defined: Why can you divide by $|G|$? Why is $\text{Sym}(\varphi)$ actually an element of $\text{Hom}_{\mathbb{K}}(V,W)$?

2. Show that $\text{Sym}$ is $\mathbb{K}$-linear.

3. Why is $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$ a subspace of $\text{Hom}_{\mathbb{K}}(V,W)?$

4. Why is $\text{Sym}(\varphi)$ an element of the above space?

5. Why is $\text{Sym}$ a projection onto $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$? That is, show that $\text{Sym}^2=\text{Sym}$ and that the image of $\text{Sym}$ is this space.Also, problem $5$ confuses me. To me, it looks like $f^n(x)$ is a polynomial of odd degree $>1$ for all $n$ (not just even). So, $f^n(x)-x$ is also an odd degree polynomial and thus has a root, which is a fixed point of $f^n$. Am I missing something? Maybe the base field is $\mathbb{Q}$ instead of $\mathbb{R}$?

 

[fresh_42]

I think Problem 2 can be done without any fiddling with commutators:

Consider the function $\{w_1,\ldots,w_n\}\to\mathbb{Z}^n, w_i\mapsto e_i.$ By the universal property of free groups, this uniquely extends to a homomorphism $F_n\to\mathbb{Z}^n.$ Note that $x\in F_n$ lies in the kernel of this map if and only if it satisfies the condition on the RHS of the equivalence (i.e. that the sums of the exponents of any fixed generator $w_i$ is zero).

Since $\mathbb{Z}^n$ is abelian, this homomorphism descends to the quotient $F_n/[F_n,F_n]\to\mathbb{Z}_n.$ We check that this map is injective by constructing a left inverse: since both groups are abelian and $\mathbb{Z}^n$ is free abelian, the function $\{e_1,\ldots,e_n\}\to F_n/[F_n,F_n]$ taking $e_i$ to the class of $w_i$ (uniquely) extends to a homomorphism $\mathbb{Z}^n\to F_n/[F_n,F_n]$ which is clearly a left inverse. (In fact, the map $F_n/[F_n,F_n]\to\mathbb{Z}^n$ is also clearly surjective so it is an isomorphism).

Now, by injectivity, $\ker(F^n\to\mathbb{Z}^n)=\ker(F^n\to F^n/[F_n,F_n]\to\mathbb{Z}^n)=[F_n,F_n].$ On the other hand, we already know that the kernel of $F^n\to\mathbb{Z}^n$ is the set of elements in $F_n$ whose exponents satisfy the given condition.

(Here, $e_i$ is the vector whose $i$-th component is $1$ and all others are zero.)

Does this count as an attempt at an answer?

The idea is correct, although you can capture all this within 4 lines. Consider for an arbitrary group $G\longrightarrow G/[G,G],$ set $G=F_n$, done.

Since you had all points in your answer I count this as solved. But I want to mention that the direct path via induction along word length is only marginally longer and still shorter than the long answer above.

 

[fresh_42]

Problem 4 is quite straightforward, but the confusing part might be figuring out exactly what is being asked. I'll list everything I think we're supposed to show and perhaps someone else would like to finish it off. I'm sure @fresh_42 will let me know if I forgot something.

1. Show that $\text{Sym}$ is well-defined: Why can you divide by $|G|$? Why is $\text{Sym}(\varphi)$ actually an element of $\text{Hom}_{\mathbb{K}}(V,W)$?

2. Show that $\text{Sym}$ is $\mathbb{K}$-linear.

3. Why is $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$ a subspace of $\text{Hom}_{\mathbb{K}}(V,W)?$

4. Why is $\text{Sym}(\varphi)$ an element of the above space?

5. Why is $\text{Sym}$ a projection onto $\text{Hom}_{\mathbb{K}}((\rho,V),(\tau,W))$? That is, show that $\text{Sym}^2=\text{Sym}$ and that the image of $\text{Sym}$ is this space.

Not quite sure, whether your point 4 equals my 'homomorphism of representations' property, but I think so.

Also, problem $5$ confuses me. To me, it looks like $f^n(x)$ is a polynomial of odd degree $>1$ for all $n$ (not just even). So, $f^n(x)-x$ is also an odd degree polynomial and thus has a root, which is a fixed point of $f^n$. Am I missing something? Maybe the base field is $\mathbb{Q}$ instead of $\mathbb{R}$?

Can happen with self-constructed problems. I was so busy to find a good polynomial, that I missed the obvious. I will correct it and make it difficult.

 

[Mayhem]

As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?

 

[fresh_42]

As for #1.

I've never done a double integral before. Do you just integrate the first integral with the respect to x and then the resulting integral with respect to y?

The integration is usually inside out according to the order of the $dx \,(1)\,dy\,(2)\,dz\,(3)$ terms.

Hint: In this case, however, nested integrals are not necessary.

 

[Office_Shredder]

I feel really dumb reading #4, because I don't even see five separate claims to prove?

 

[kshitij]

Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

edit: In the above answer, $a^2+b^2\neq 0$

 

[fresh_42]

Problem #15

$$\begin{align}
x+y&=az\\
x-y&=bz\\
x^2+y^2&=cz
\end{align}$$

Squaring equations (1) & (2) and adding,
$$\begin{align}
2cz=(a^2+b^2)z^2\nonumber\\
z=0, \frac {2c} {a^2+b^2}\nonumber
\end{align}$$

Adding & subtracting equations (1) & (2) respectively,
$$\begin{align}
2x=z(a+b)\nonumber\\
2y=z(a-b)\nonumber
\end{align}$$
putting the value of z
$$(x,y,z) = (0,0,0) \space or \left( \frac {c(a+b)} {a^2+b^2},\frac {c(a-b)} {a^2+b^2},\frac {2c} {a^2+b^2} \right)$$

Why does $x=y=0$ imply $z=0$?

 

[kshitij]

Why does x=y=0 imply z=0?

I used that z=0 imply x=y=0

as we have $2cz=(a^2+b^2)z^2$ from here we get one value of z as zero and substituting that we get x=y=0

 

[fresh_42]

I used that z=0 imply x=y=0

as we have $2cz=(a^2+b^2)z^2$ from here we get one value of z as zero and substituting that we get x=y=0

Sure, but what if $z\neq 0$? You missed a possibility.

 

[kshitij]

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)

 

[kshitij]

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)

 

[fresh_42]

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.

 

[kshitij]

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?

 

[kshitij]

Your first post was already correct, except one special case.

Is a=b=0 the special case you were talking about here?

 

[fresh_42]

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.

 

[kshitij]

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

 

[fresh_42]

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 $z$.

 

[kshitij]

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.

 

[fishturtle1]

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. ?

 

[fresh_42]

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?

 

[fishturtle1]

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 $W$.

 

[fresh_42]

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.

 

[fishturtle1]

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 $W$?

 

[fresh_42]

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.

 

[fishturtle1]

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!

 

[fishturtle1]

Spoiler: problem 4

\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.
[]

 

[fresh_42]

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":

• $\tau(h)\circ\operatorname{Sym}(\varphi)=\operatorname{Sym}(\varphi)\circ \rho(h)$
• $\operatorname{Sym}^2(\varphi )=\operatorname{Sym}(\varphi )$

 

[fresh_42]

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?

 

[fishturtle1]

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.