I would like to know if the following conjecture is correct and if so what's a good citation for its proof.


Let $\mathsf{E}$ be the category of euclidean vector spaces, i.e. objects are finite-dimensional real vector spaces endowed with a scalar product and morphisms are isometries. The tensor powers $(-)^{\otimes a}$ are functors $\mathsf{E}\to\mathsf{Vect}$. There are some obvious natural transformations between these functors. For example taking scalar products of any two tensor factors gives $\binom{a}{2}$ many natural transformations $(-)^{\otimes a} \to (-)^{\otimes(a-2)}$. Permuting tensor functors gives $a!$ many natural transformations $(-)^{\otimes a}\to(-)^{\otimes a}$. Of course linear combinations and compositions of these are also natural transformations.


The conjecture is that these are essentially all there is to have. More precisely I think that the following statement is true:



> 
> Let $a,b\in\mathbb{N}$ be fixed. The only natural transformations $(-)^{\otimes a} \to (-)^{\otimes b}$ are either zero if $b-a \notin 2\mathbb{N}$ or compositions of an element of $\mathbb{R}[S\_a]$ (acting by permuting the tensor factors) and $\frac{b-a}{2}$ many scalar products if $b-a\in 2\mathbb{N}$
> 
> 
> 


Considering Qiaochu Yuan's comment, a second question is also interesting to ask: If one considers $Iso(\mathsf{E})$ (i.e. the category with only bijective isometries as morphisms) instead of $\mathsf{E}$, how can we characterise the natural transformations between the analogous functors then? The Casimir element $\Omega\_V:=\sum\_i b\_i\otimes b\_i$ is independent of the choice of the basis $b\_1,\ldots,b\_n$ of $V$ and therefore gives rise to different natural transformations like $\mathbb{R}\to V^{\otimes 2}, 1\mapsto\Omega\_V$. I'd guess that that's essentially it and every natural transformation is composed of $\mathbb{R}[S\_a]$, some number of traces and some number of insertions of $\Omega$.



Characterising natural transformations between tensor functors
The second question has a well known, affirmative answer from invariant theory.


Note that $O(V)$ acts on every tensor power $V^{\otimes k}$ and by conjugation also on $Hom(V^{\otimes a},V^{\otimes b})$ and the homomorphisms commuting with $O(V)$ are exactly the fixed points of this conjugation action. We have a natural isomorphism
$$Hom(V^{\otimes a},V^{\otimes b}) \cong (V^{\otimes a})^\ast \otimes V^{\otimes b} \cong (V^\ast)^{\otimes a}\otimes V^{\otimes b}$$
The scalar product gives us an isomorphism $V\cong V^\ast$ which translates this into
$$Hom(V^{\otimes a},V^{\otimes b}) \cong V^{\otimes a}\otimes V^{\otimes b} = V^{\otimes(a+b)}$$
Explicitly this last one is given by
$$x\_1\otimes\cdots\otimes x\_a \otimes y\_1\otimes \cdots \otimes y\_b \mapsto \left(v\_1\otimes\cdots\otimes v\_a \mapsto \prod\_{i=1}^a \langle x\_i,v\_i\rangle y\_1\otimes\cdots\otimes y\_b\right)$$
All of these are natural w.r.t. isometries so that our problem translates into finding fixed points on $V^{\otimes k}$ for every $k\in\mathbb{N}$. Since $v\mapsto -v$ is an isometry, there are no non-zero fixed points for uneven $k$. So we are left with the case where $a+b$ is even.


For the special case $a=b$, we want the centraliser of $\rho(O(V))$ inside $End(V^{\otimes a})$ where $\rho$ is the representation $O(V)\to End(V^{\otimes a}), \phi\mapsto\phi^{\otimes a}$. This centraliser turns out to be well-known: It is $\psi(B\_a(d))$ where $d:=\dim(V)$, $B\_a(d)$ is the so-called [Brauer algebra](https://en.wikipedia.org/wiki/Brauer_algebra) and $\psi: B\_a(d) \to End(V^{\otimes a})$ is its natural representation. This is the analogue of Schur-Weyl-duality for orthogonal groups.


The Brauer algebra has a basis indexed by perfect pairings $D$ of the numbers $1,2,\ldots,2a$ each of which describes a combination of permutations (when a number in $\{1,\ldots,a\}$ is paired with a number in $\{a+1,\ldots,2a\}$) and the map $v\_1\otimes v\_2 \mapsto \langle v\_1,v\_2\rangle \Omega$, where $\Omega:=\sum\_{i=1}^{\dim(V)} e\_i\otimes e\_i$ is the Casimir element.


If we follow the isomorphisms above, we find that these basis elements give us the following spanning set of fixed points in $(V^{\otimes 2a})^{O(V)}$:
$$\Omega\_D := \sum\_{\substack{1\leq i\_1,\ldots,i\_{2a}\leq \dim(V) \\ \lbrace x,y\rbrace\in D \implies i\_x=i\_y}} e\_{i\_1} \otimes e\_{i\_2} \otimes \cdots \otimes e\_{i\_{2a}}$$
where $D$ runs over all perfect pairings of the numbers $1,2,\ldots,2a$.


Now the fixed points do not care that arrived at the number $2a$ by considering $a=b$. Therefore this is also the answer for the general case where $a+b$ is any even number. If we follow all the isomorphisms backward, we find that these fixed points $\Omega\_D\in V^{\otimes(a+b)}$ represent the maps $\tau^D: V^{\otimes a}\to V^{\otimes b}$ that come from permuting some tensor factors (=numbers in $\{1,\ldots,a\}$ get paired with numbers in $\{a+1,\dots,a+b\}$), contracting some (=two numbers in $\{1,\ldots,a\}$ get paired with each other), and introducing Casimir elements in some factors of the result (=two numbers in $\{a+1,\ldots,a+b\}$ get paired with each other) which is what we wanted to prove.




---


To prove the original conjecture, we have to find all the linear combinations
$$\tau = \sum\_D \alpha\_D \tau^D$$
which are natural not only w.r.t to bijective isometries but w.r.t. all isometries. Our theorem is proven if we can prove that
$$D\text{ contains a pairing }\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\} \implies \alpha\_D=0$$
Because then all introductions of Casimir elements are eliminated and only traces and reordering of tensor factors remain. In particular it would follow that $a\geq b$ must hold if $\tau\neq 0$. That together with the earlier parity argument would prove our theorem.


Step 1.: 
Consider a vector space $V$ of dimension $d$ with an orthonormal basis $e\_1,\ldots,e\_d$ and define for any assignment of indices $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ the tensor $v\_j\in V^{\otimes a}$ as
$$v\_j := e\_{j(1)} \otimes e\_{j(2)} \otimes \cdots \otimes e\_{j(a)}$$


If we have two such assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$, then we can compute the scalar product $\langle{\tau\_V(v\_j),v\_k}\rangle$ as follows:
\begin{align\*}
 \langle{\tau\_V(v\_j),v\_k}\rangle &= \sum\_D \alpha\_D \langle{\tau\_V^D(v\_j),v\_k}\rangle \\
 &= \sum\_{D,i} \alpha\_D \prod\_{s=1}^a \langle{e\_{i(s)},e\_{j(s)}}\rangle \langle{e\_{i(a+1)}\otimes\cdots\otimes e\_{i(a+b)}, e\_{k(1)}\otimes\cdots\otimes e\_{k(b)}} \rangle \\
 &= \sum\_{D,i} \alpha\_D \prod\_{s=1}^a \langle{e\_{i(s)},e\_{j(s)}}\rangle \prod\_{t=1}^b \langle{e\_{i(a+t)}, e\_{k(t)}}\rangle
\end{align\*}


where - as above - $i$ runs over all indexing functions $i:\{1,2,\ldots,a+b\}\to\{1,\ldots,d\}$ such that
$$\{x,y\}\in D \implies i(x)=i(y).\tag{\*}$$


Step 2: We prove the following lemma



> 
> If $d\geq \frac{a+b}{2}$, then for every perfect pairing $D$, there are assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$ such that
>  $$\langle{\tau\_V(v\_j),v\_k}\rangle = \alpha\_D$$
>  If moreover $D$ contains a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$, then it is possible to find such $j$ and $k$ with $\forall 1\leq x\leq a: j(x) \leq m-1$.
> 
> 
> 


Proof of the Lemma:
To see this, note that the factors in both products in the above expression for the scalar product are either one or zero so that if the whole summand is non-zero, then $\forall s: i(s) = j(s)$ and $\forall t: i(a+t) = k(t)$ must hold, i.e. there can only be one $i$: The one that agrees with $j$ on $\{1,\ldots,a\}$ and with $k$ on $\{a+1,\ldots,a+b\}$. Therefore most summands in the expression are zero anyway and can be ignored. The sum reduces to
$$\langle{\tau\_V(v\_j),v\_k}\rangle = \sum\_{D} \alpha\_D \prod\_{s=1}^a \langle{e\_{i(s)},e\_{j(s)}}\rangle \prod\_{t=1}^b \langle{e\_{i(a+t)}, e\_{k(t)}}\rangle$$
where $i$ is the one function that agrees with $j$ in the lower and with $k$ in the upper part of $\{1,2,\ldots,a,a+1,\ldots,a+b\}$.


Additionally, there are only a limited number of pairings $D$ that are compatible with this indexing function $i$, because $i$ also has to satisfy the condition $(\*)$.
Now let our given pairing $D$ be $\Big\{\{x\_1,y\_1\}, \{x\_2,y\_2\}, \ldots, \{x\_m,y\_m\}\Big\}$. Then if we choose $j$ and $k$ such that $i(x\_1)=i(y\_1) = 1$, $i(x\_2)=i(y\_2)=2$, ..., $i(x\_m)=i(y\_m)=m$, then there is only a single pairing that satisfies $(\ast)$ for this indexing function $i$, namely the pairing $D$ we started with.


All summands other than the $D$-th are therefore zero and the whole sum simplifies to
$$\langle{\tau\_V(v\_j),v\_k}\rangle = \alpha\_D \prod\_{s=1}^a \langle{e\_{i(s)},e\_{j(s)}}\rangle \prod\_{t=1}^b \langle{e\_{i(a+t)}, e\_{k(t)}}\rangle = \alpha\_D\cdot 1$$


We have complete freedom which numbering of the pairs in $D$ we choose to define $i$. If $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$ is one of the pairs in $D$, we can choose our ordering such that $x\_m=\alpha$ and $y\_m=\beta$. Therefore the numbers $1,2,\ldots,a$ get mapped to something other than $m$ by $i$. That proves the lemma.


Now back to the proof of the theorem.


Step 3. Let's focus on a specific example: Choose $V$ to be large enough that $d\geq \frac{a+b}{2}$ and set $U:=span(e\_1,\ldots,e\_{m-1})$. The embedding $\phi: U\to V, v\mapsto v$ is an isometry so that our natural transformation $\tau$ must satisfy
$$\forall v\in U^{\otimes a}: \tau\_U(v) = \tau\_V(v)$$
Now in $\tau\_U(v)$ all the sums only have indices running up to $\dim(U)=m-1$ whereas in $\tau\_V(v)$ the same indices run up to $d$. On the right hand side however some summands containing $e\_m,e\_{m+1},\ldots,e\_d$ may appear.


Consider the linear combination $\tau = \sum\_D \alpha\_D \tau^D$ we started with and let $D$ be one of the perfect pairings which contain a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$. By the lemma just proven, there is an indexing function $i$ and its two parts $j$ and $k$ such that $j$ takes only values inside $\{1,\ldots,m-1\}$ and $\langle{\tau\_V(v\_j),v\_k}\rangle = \alpha\_D$.


Moreover, because $j$ takes values in $\{1,\ldots,m-1\}$, $v\_j$ must be in $U^{\otimes a}$ and $v\_k$ contains $e\_m,e\_{m+1},\ldots,e\_d \in U^\perp$ in the $\alpha$-th (and $\beta$-th too) tensor factor. We conclude
$$\alpha\_D = \langle{\tau\_V(v\_j),v\_k}\rangle = \langle{\tau\_U(v\_j),v\_k}\rangle = 0$$
and that proves the theorem.


