An iterated tensor product integral

Question

In "Differential equations driven by rough paths" (Terry Lyons, et al) section 1.4.2 it's claimed that the symmetric part of the tensor:

$\int_{0 \le u_1 \le \cdots \le u_j \le t} \mathrm{d}X_{u_1} \otimes \cdots \otimes \mathrm{d}X_{u_j}$

is exactly $\frac{1}{j!}(X_t - X_0)^{\otimes j}$. It is assumed that $X:[0,T] \to V$ is a Lipschitz path in some finite dimensional vector space $V$.

Is there a simple way to establish this fact? Also I would be grateful for any reference on this type of integrals involving tensors since I've never encountered them before.

Further background

Integration with respect to $X$ is defined for any continuous $f: [0,T] \to L(V,W)$ ($W$ a finite dimensional vector space and $L(V,W)$ the space of linear maps from $V$ to $W$) as:

$\int f_s \mathrm{d}X_s = \lim \sum_{i = 0}^{r -1} f_{t_i}(X_{t_{i+1}} - X_{t_i})$

where the limit is taken over partitions $t_0 = 0 \le t_1 \le \cdots \le t_r = T$ whose largest interval is decreasing to $0$ in length. The result belongs to the space $W$.

Hence, to make the above iterated integral fit into the definition (and give a result in $V^{\otimes j}$) it seems to me that one must consider each element $w_1 \otimes \cdots \otimes w_i \in V^{\otimes i}$ as an element of $L(V, V^{\otimes i+1})$ via the map $v \mapsto w_1 \otimes \cdots \otimes w_i \otimes v$. However after doing this I still have trouble establishing the identity.

As an example, with the above interpretation I get:

$\int_{0 \le s \le t \le T} \mathrm{d}X_s \otimes \mathrm{d}X_t = \frac{1}{2}T^2 v_1\otimes v_1 + \frac{2}{3}T^3 v_1\otimes v_2 + \frac{1}{3}T^3 v_2 \otimes v_1 + \frac{1}{2}T^4 v_2 \otimes v_2$

for $X_t = t v_1 + t^2 v_2$ in a $2$-dimensional vector space with basis $(v_1,v_2)$. This example shows that the result of the integral itself may be non-symmetric.

Answer 1

Let's just view the function $\dot{X}(t) = \frac{d}{dt} X_t$ as a bounded function taking values in the vector space $V$. The notation $dX_{u_1}$ means $\dot{X}(u_1) du_1$ Then $dX_{u_1} \otimes \cdots \otimes dX_{u_k} = \dot{X}(u_1) \otimes \cdots \otimes \dot{X}(u_k) du_1 \cdots du_k$, should be regarded as a bounded function (or measure) on $[0,t]^k$ taking values in the vector space $V^{\otimes k}$.

Because the operation of projecting to a symmetric part is a linear operation from $V^{\otimes k}$ to itself, you can take it inside of the integral. Let's call this symmetric part $dX_{u_1} \cdots dX_{u_k}$. I'll use the symbol $u \cdot v$ to denote the symmetric product of of $u, v \in V$ -- that is, $u \cdot u \cdot u = u \otimes u \otimes u$, and the general product is defined by the polarization identity. For example $u \cdot v = ( u \otimes v + v \otimes u) / 2$. The resulting multiplication is commutative.

Therefore your integral is equal to

$\int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k}$

Because the symmetric product is commutative, for any permutation $\sigma$ of $\{ 1 \ldots k\}$ you get the same value by integrating any of the permuted regions

$ \int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{0 \leq u_{\sigma(1)} \leq \ldots \leq u_{\sigma(k)} \leq t} dX_{u_1} \cdots dX_{u_k}$

Observe also that the region $0 \leq u_1 \leq \ldots \leq u_k \leq t$ is a fundamental domain for the action of $S_k$ on the cube $[0,t]$ -- that is, you can get the whole cube by permuting the variables $u_i$, and none of these regions overlap. Therefore, since there are $k!$ such permutations, we sum over permutations to conclude

$ k! \int_{0 \leq u_1 \leq \ldots \leq u_k \leq t} dX_{u_1} \cdots dX_{u_k} = \int_{[0,t]^k} dX_{u_1} \cdots dX_{u_k}$

But, by Fubini and the multilinearity of the symmetric product, the integral on the right is just $(X_t - X_0)^k$.

I should remark that this same proof in an even simpler setting also produces the $\frac{1}{k!}$ that appears when you prove the Taylor expansion through iterative use of the fundamental theorem of calculus.