# Characterising natural transformations between tensor functors

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 $$\Omega$$.

• There is also the dual of the scalar product, which raises degree. – Qiaochu Yuan Oct 8 '19 at 20:11
• You mean the map $\mathbb{R}\to V\otimes V, 1\mapsto \sum_i b_i\otimes b_i$ where $\lbrace b_i \mid i=1...\dim(V)\rbrace$ is an orthonormal basis of $V$ ? I really missed that map. However, I don't think it is natural because it isn't compatible with the inclusion of subspaces. – Johannes Hahn Oct 8 '19 at 20:34

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 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.