# The K-ring of $\mathrm{E}_{6}/\mathrm{Spin}(10)$

Sudeep Podder  Department of Mathematics, Indian Institute of Technology
Madras, Chennai 600036, India<EMAIL_ADDRESS>and  Parameswaran
Sankaran  Chennai Mathematical Institute, SIPCOT IT Park, Siruseri,
Kelambakkam, 603103, India<EMAIL_ADDRESS>
(Date: August 28, 2024)

###### Abstract.

Let $\mathrm{E}_{6}$ denote the simply-connected compact exceptional Lie group
of rank $6$. The Lie group $\mathrm{Spin}(10)$ naturally embeds in
$\mathrm{E}_{6}$, corresponding to the inclusion of the Dynkin diagrams. We
determine the $K$-ring of the coset space $\mathrm{E}_{6}/\mathrm{Spin}(10)$.
We identify the class of the tangent bundle of
$\mathrm{E}_{6}/\mathrm{Spin}(10)$ in $KO(\mathrm{E}_{6}/\mathrm{Spin}(10))$.
As an application we show that $\mathrm{E}_{6}/\mathrm{Spin}(10)$ can be
immersed in the Euclidean space $\mathbb{R}^{53}.$

###### Key words and phrases:

Quotients of the exceptional Lie group $\mathrm{E}_{6}$, $K$-theory, Hodgkin
spectral sequence.

###### 2010 Mathematics Subject Classification:

Primary: 55N15; Secondary: 19L99

##  1\. Introduction

The purpose of this article is to determine the ring structure of the complex
$K$-theory of the compact homogeneous space
$M=\mathrm{E}_{6}/\mathrm{Spin}(10)$ where $\mathrm{Spin}(10)$ is embedded in
the compact simply-connected Lie group $\mathrm{E}_{6}$ obtained from the
embedding of the Dynkin diagram of $\mathrm{Spin}(10)$ into that of
$\mathrm{E}_{6}$. There are two embeddings of the Dynkin diagram of
$\mathrm{Spin}(10)$ corresponding to omitting any of the two end roots of the
‘long arms’ of the Dynkin diagram of $\mathrm{E}_{6}$. They are mapped to one
another by a diagram automorphism of $\mathrm{E}_{6}$, so the corresponding
embeddings of $\mathrm{Spin}(10)$ are mapped to one another by an outer
automorphism of $\mathrm{E}_{6}$. So the corresponding homogeneous spaces are
diffeomorphic. The manifold $M$ is the principal $\mathbb{S}^{1}$ bundle over
$X=\mathrm{E}_{6}/(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})$, which is a
Hermitian symmetric space. It is known that $M\times\mathbb{S}^{1}$ admits the
structure of the total space of a complex analytic bundle with fibre an
elliptic curve. See [21]. However, $M\times\mathbb{S}^{1}$ does not admit a
Kähler structure since its first Betti number is $1$. The $K$-ring of $X$ is
known. Indeed if $Y$ is any homogeneous space $G/H$ where $G$ is a compact
simply connected Lie group and $H$, a closed connected Lie subgroup of $G$
having the same rank as $G$, then it is known that
$K^{0}(Y)=RH\otimes_{RG}\mathbb{Z}$. Here $RG$ is the complex representation
ring of $G$, $RH$ is an $RG$-module via the restriction homomorphism $RG\to
RH$, and $\mathbb{Z}$ is the $RG$-module via the augmentation
$RG\to\mathbb{Z}$ that maps $[V]$ to $\dim V$ for any complex representation
$V$ of $G$. See [5] and [15].

We now state the main results of this note. Denote by $\lambda_{1}$ the
standard representation of $\mathrm{SO}(10)$ on $\mathbb{C}^{10}$, viewed as a
representation of $\mathrm{Spin}(10)$ via the double covering
$\mathrm{Spin}(10)\to\mathrm{SO}(10)$. We shall denote by the same symbol
$\lambda_{1}$ the $\alpha$-construction on $\lambda_{1}$. (Thus the total
space $E(\lambda_{1})$ equals
$\mathrm{E}_{6}\times_{\mathrm{Spin}(10)}\mathbb{C}^{10}$ with projection
$E(\lambda_{1})\to\mathrm{E}_{6}/\mathrm{Spin}(10)$ defined as $[x,v]\mapsto
x\mathrm{Spin}(10)\leavevmode\nobreak\ \forall
x\in\mathrm{E}_{6},v\in\mathbb{C}^{10}.$)

######  Theorem 1.1.

$K^{0}(\mathrm{E}_{6}/\mathrm{Spin}(10))\cong\mathbb{Z}[u]/\langle
u^{3}\rangle$ where $u=\lambda_{1}-10$. The
$K^{1}(\mathrm{E}_{6}/\mathrm{Spin}(10))$ is a free
$K^{0}(\mathrm{E}_{6}/\mathrm{Spin}(10))$-module of rank $1$.

We identify the class of the tangent bundle of
$\mathrm{E}_{6}/\mathrm{Spin}(10)$ in its $KO$-ring and obtain the following.

######  Theorem 1.2.

The manifold $\mathrm{E}_{6}/\mathrm{Spin}(10)$ can be immersed in
$\mathbb{R}^{53}$ but not in $\mathbb{R}^{40}.$

The non-immersion result is essentially due to Singhof and Wemmer [20, §3].

Our proof of Theorem 1.1 involves use of the Hodgkin spectral sequence [13],
which will be recalled briefly in §5.1. As a crucial computational aid, we
will also need the ‘change of ring theorem’, namely [7, Theorem 6.1, Chapter
16]. The $E_{2}$-page of the Hodgkin spectral sequence involves the
representation ring of $\mathrm{E}_{6}$ and the structure of
$R\mathrm{Spin}(10)$ as a module over $R\mathrm{E}_{6}$ via the restriction
homomorphism $R\mathrm{E}_{6}\to R\mathrm{Spin}(10)$. They will be described
in §4. For the sake of completeness, we describe $\mathrm{E}_{6}$ as a
subgroup of the compact Lie group $\mathrm{E}_{8}$. This will also help us
describe the embedding of $\mathrm{Spin}(10)$ in $\mathrm{E}_{6}$. The main
theorems are proved in §5.2 and §5.4.

We have chosen to study the $K$-theory of $\mathrm{E}_{6}/\mathrm{Spin}(10)$
since it is the smallest dimensional quotient $\mathrm{E}_{6}/H$ where $H$ is
connected and of rank less than $6$. Although the tools employed here would be
applicable in full generality, the computation of the $E_{2}$-page
$\mathrm{Tor}^{*}_{R\mathrm{E}_{6}}(RH,\mathbb{Z})$ appears to be rather
unwieldy.

We assume familiarity with the basic notions of finite dimensional
representations of compact Lie groups as can be found in [1], [8], the
Clifford algebra and the spin representations (as in [14]), as well as the
description of the exceptional Lie group $\mathrm{E}_{6}$. A good reference
for the construction of the exceptional simple Lie groups is the book [3].

##  2\. $\mathrm{E}_{6}$ as a subgroup of $\mathrm{E}_{8}$

We begin by describing the compact simply connected exceptional Lie group
$\mathrm{E}_{6}$ as a subgroup of the compact exceptional Lie group
$\mathrm{E}_{8}$. We follow [3] closely. This will allow us to describe the
‘standard’ maximal torus of $\mathrm{E}_{6}$ as a subgroup of the ‘standard’
maximal torus of $\mathrm{E}_{8}$. This will be needed to carry out our
description of the restriction homomorphism $R\mathrm{E}_{6}\to
R\mathrm{Spin}(10)$.

The subgroup $\mathrm{E}_{6}$ of $\mathrm{E}_{8}$ is the centralizer of a copy
of $\mathrm{SU}(3)$ contained in $\mathrm{E}_{8}$. In order to describe this
subgroup (to be identified with $\mathrm{SU}(3)$) and also to describe the
fundamental representations of $\mathrm{E}_{6}$, we need the description of
the Lie algebra $Lie(\mathrm{E}_{8})$. To fix notations, we need to label of
nodes of the Dynkin diagram of $\mathrm{E}_{8}$. We follow the labelling
convention as in [3], given below.

$1\ \ \ \ \ \ \ \ \ \ \ 2\ \ \ \ \ \ \ \ \ \ \ \ 4\ \ \ \ \ \ \ \ \ \ \ \ 3\ \
\ \ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ \ \ \ \ \ \ 7\ \ \ \ \ \ \ \ \ \ \ 8$$5$
Figure 1. Dynkin diagram of $\mathrm{E}_{8}.$

Consider the half-spin representations $\Delta^{\pm}_{8}$ of
$\mathrm{Spin}(16)$. They are complexifications of the real representations
$\Delta^{\pm}_{8,\mathbb{R}}$. It turns out that
$Lie(\mathrm{Spin}(16))\oplus\Delta^{+}_{8,\mathbb{R}}$ has the structure of a
compact Lie algebra, extending the Lie algebra structure on
$Lie(\mathrm{Spin}(16))$ and the $Lie(\mathrm{Spin}(16))$-module structure of
$\Delta_{8,\mathbb{R}}^{+}.$ This is the Lie algebra of $\mathrm{E}_{8}$ and
the group $\mathrm{E}_{8}$ is the group of automorphisms of
$Lie(\mathrm{Spin}(16))\oplus\Delta^{+}_{8,\mathbb{R}}$, which is viewed as
the adjoint action. See [3, Chapter 7]. Evidently, we have a homomorphism
$\mathrm{Spin}(16)\to\mathrm{E}_{8}$ since $\mathrm{Spin}(16)$ is simply
connected. Its image is the semi-spin group
$\mathrm{Spin}(16)/\mathbb{Z}_{2}\subset\mathrm{E}_{8}$, where
$\mathbb{Z}_{2}$ is generated by the central element $\zeta_{8}:=e_{1}\cdots
e_{16}\in\mathrm{Spin}(16)$. Here the $e_{j}$ denotes the standard basis
vectors of $\mathbb{R}^{16}$ and are viewed as the Clifford algebra units. The
reader is referred to Husemoller’s book [14, Chapter 11] for basic information
on Clifford algebras, spin groups and the (half-) spin representations.

Consider the canonical monomorphism $\mathrm{SU}(3)\to\mathrm{SO}(6)$. Since
$\mathrm{SU}(3)$ is simply connected, this monomorphism lifts to a map
$\psi:\mathrm{SU}(3)\to\mathrm{Spin}(6)$. This, followed by the obvious
homomorphism $\mathrm{Spin}(6)\to\mathrm{Spin}(16)\to\mathrm{E}_{8}$, yields a
homomorphism $\mathrm{SU}(3)\to\mathrm{E}_{8}$. It is, in fact, a monomorphism
and corresponds to the inclusion of the Dynkin diagram of $\mathrm{SU}(3)$,
regarded as the edge joining the last two nodes of the long arm of the Dynkin
diagram of $\mathrm{E}_{8}$. These are labelled $7$ and $8$ in Figure 1.

${\widetilde{\rm T}_{3}}$${\widetilde{\rm T}_{5}\times\widetilde{\rm
T}_{3}}$${\widetilde{\rm T}_{8}}$${\rm
T^{8}}$${\mathrm{SU}(3)}$${\mathrm{Spin}(6)}$${\mathrm{Spin}(10)\times\mathrm{Spin}(6)}$${\mathrm{Spin}(16)}$${\mathrm{E}_{8}}$${\mathrm{U}(3)}$${\mathrm{SO}(6)}$$\scriptstyle{\psi}$$\scriptstyle{\pi}$
Figure 2.

The exceptional group $\mathrm{E}_{6}$ is then the centralizer of the image of
$\mathrm{SU}(3)$ in $\mathrm{E}_{8}$. It is a simply connected group, and its
centre is a cyclic group of order $3$. The Dynkin diagram of $\mathrm{E}_{6}$
is obtained by omitting the nodes $7$ and $8$. We shall describe the
‘standard’ maximal torus $\rm T^{6}$ of $\mathrm{E}_{6}$ and describe $R\rm
T^{6}$ and $R\mathrm{E}_{6}$ using the description of the ‘standard’ maximal
torus $\rm T^{8}$ of $\mathrm{E}_{8}$. This will enable us to determine the
restriction homomorphism $R\mathrm{E}_{6}\to RH$ when $H=\mathrm{Spin}(10).$

###  2.1. The maximal tori of $\mathrm{E}_{6}$ and $\mathrm{Spin}(10)$

Let $\mathbb{T}_{k}=(\mathbb{S}^{1})^{k}=\mathbb{R}^{k}/\mathbb{Z}^{k}$, the
standard $k$-dimensional torus. Let $\rm T_{k}$ be the subgroup
$\mathrm{SO}(2)\times\cdots\times\mathrm{SO}(2)\subset\mathrm{SO}(2k)$. Thus
$\rm T_{k}$ is generated by proper rotations on the plane
$\mathbb{R}e_{2j-1}+\mathbb{R}e_{2j}\subset\mathbb{R}^{2k},1\leq j\leq k.$
Then $\widetilde{\rm T}_{k}=\pi^{-1}(\rm T_{k})$ is a maximal torus of
$\mathrm{Spin}(2k)$ where $\pi:\mathrm{Spin}(2k)\to\mathrm{SO}(2k)$ is the
double covering projection. One has the homomorphism
$\bar{\omega}_{k}:\mathbb{T}_{k}\to\rm T_{k}$ defined as
$(\theta_{1},\ldots,\theta_{k})\mapsto
D(2\theta_{1},\ldots,2\theta_{k}):=\prod_{1\leq j\leq k}D_{j}(2\theta_{j})$
where $D_{j}(t)(e_{2j-1})=\cos(2\pi t)e_{2j-1}-\sin(2\pi
t)e_{2j},D_{j}(t)(e_{2j})=\sin(2\pi t)e_{2j-1}+\cos(2\pi t)e_{2j}$ and
$D_{j}(t)(e_{k})=e_{k}$ if $k\neq 2j-1,2j$.

Then $\bar{\omega}_{k}$ lifts to a unique homomorphism
$\omega_{k}:\mathbb{T}_{k}\to\widetilde{\rm T}_{k}$ so that
$\bar{\omega}_{k}=\pi\circ\omega_{k}$. It is readily seen that
$\ker(\bar{\omega}_{k})=\\{1,-1\\}^{k}\subset(\mathbb{S}^{1})^{k}$ and that
$\ker(\omega_{k})=\\{(\varepsilon_{1},\ldots,\varepsilon_{k})\in\ker(\bar{\omega})\mid\prod\varepsilon_{j}=1\\}$.
The complex representation rings of $\mathbb{T}_{k},\widetilde{\rm T}_{k}$,
and $\rm T_{k}$ are related as follows: Let $u_{j}$ be the one dimension
(complex) representation of $\mathbb{T}_{k}$ corresponding to the $j$th
projection $u_{j}:\mathbb{T}_{k}\to\mathbb{S}^{1}=\rm U(1)$ viewed as a
character. Then $R\mathbb{T}_{k}=\mathbb{Z}[u_{1}^{\pm 1},\ldots,u_{k}^{\pm
1}]$, $R\rm T_{k}=\mathbb{Z}[u_{1}^{\pm 2},\ldots,u_{k}^{\pm 2}]\subset
R\mathbb{T}_{k}$ and $R\widetilde{\rm T}_{k}=\mathbb{Z}[u_{1}^{\pm
2},\ldots,u_{k}^{\pm 2},u_{1}\ldots u_{k}]\subset R\mathbb{T}_{k}$. The double
covering $\widetilde{\rm T}_{k}\to\rm T_{k}$ induces the inclusion $R{\rm
T}_{k}\subset R\widetilde{\rm T}_{k}$. We shall abbreviate $\omega_{k}$ and
$\bar{\omega}_{k}$ to $\omega$ and $\bar{\omega}$ when $k$ is clear from the
context. Our main interest is when $k=5,8$.

We have the diagonal embedding
$j:\mathbb{S}^{1}\hookrightarrow\mathbb{T}_{3}$. Define
$\delta:\mathbb{S}^{1}\to\mathrm{Spin}(6)$ to be $z\mapsto\omega_{3}(j(z))$.
It is readily seen that $\delta$ is a monomorphism. Then
$S:=\delta(\mathbb{S}^{1})$ is contained in $\widetilde{\rm T}_{3}$ and
centralizes the image of $\mathrm{SU}(3)$ under $\psi$. So we obtain a
homomorphism $S\times\mathrm{SU}(3)\to\mathrm{Spin}(6)$ defined as
$(z,A)\mapsto z\cdot\psi(A)$

We have the natural homomorphism
$\mathrm{Spin}(10)\times\mathrm{Spin}(6)\to\mathrm{Spin}(16)$ and composing
with the homomorphism $\mathrm{Spin}(16)\to\mathrm{E}_{8}$ we obtain a
homomorphism $\Psi$ defined as the composition

(1)
$\Psi:\mathrm{Spin}(10)\times\mathbb{S}^{1}\times\mathrm{SU}(3)\stackrel{{\scriptstyle
h}}{{\to}}\mathrm{Spin}(10)\times\mathrm{Spin}(6)\to\mathrm{Spin}(16)\to\mathrm{E}_{8}$

where $h$ is defined as $(v,z,A)\mapsto(v,\delta(z)\cdot\psi(A))$. The
homomorphism maps $\mathrm{SU}(3)$ monomorphically into $\mathrm{E}_{8}$ and
the image $\Psi(\mathrm{SU}(3))$ corresponds to the Lie subalgebra of
$\mathrm{E}_{8}$ whose Dynkin diagram is the edge of Dynkin diagram of
$\mathrm{E}_{8}$ joining the nodes $7,8.$ The subgroup
$\mathrm{E}_{6}\subset\mathrm{E}_{8}$ is, by definition, the centralizer of
this copy of $\mathrm{SU}(3)$ in $\mathrm{E}_{8}$. Clearly,
$\Psi(\mathrm{Spin}(10)\times\mathbb{S}^{1})$ is contained in
$\mathrm{E}_{6}$. It can be seen that
$\Psi|_{\mathrm{Spin}(10)\times\mathbb{S}^{1}}$ is a local isomorphism whose
kernel is the cyclic subgroup $Z$ of order $4$, generated by $(e_{1}\cdots
e_{10},i)\in\mathrm{Spin}(10)\times\mathbb{S}^{1}$. Therefore
$\Psi(\mathrm{Spin}(10)\times\mathbb{S}^{1})=\mathrm{Spin}(10)\times_{Z}\mathbb{S}^{1}$.
The group $\mathrm{Spin}(10)$ is mapped injectively into $\mathrm{E}_{6}$ by
$\Psi.$ The subgroup $\rm
T^{6}:=\Psi(\tilde{T}_{5}\times\mathbb{S}^{1})\subset\mathrm{E}_{6}$ is a
maximal torus of $\mathrm{E}_{6}$ which contains the maximal torus
$\widetilde{\rm T}_{5}$ of $\mathrm{Spin}(10).$ These are the standard maximal
tori of $\mathrm{E}_{6}$ and $\mathrm{Spin}(10).$

##  3\. Representation Ring of $\mathrm{E}_{6}$.

Recall that $\mathbb{T}_{8}=(\mathbb{S}^{1})^{8}$ is the standard torus of
dimension $8$, whose representation ring is the Laurent polynomial ring
$R\mathbb{T}_{8}=\mathbb{Z}[u_{1}^{\pm 1},\ldots,u_{8}^{\pm 1}]$. Here,
$u_{j}$ is the class of the $1$-dimensional representation with character the
$j$th coordinate projection
$u_{j}:\mathbb{T}_{8}\to\mathbb{S}^{1},\leavevmode\nobreak\ 1\leq j\leq 8.$
Recall the covering projection $\mathbb{T}_{8}\to\widetilde{\rm T}_{8}$, where
$\widetilde{\rm T}_{8}$ is the ‘standard’ maximal torus of
$\mathrm{Spin}(16).$ The representation ring of $\widetilde{\rm T}_{8}$ is
$\mathbb{Z}[u_{1}^{\pm 2},\ldots,u_{8}^{\pm 2},u_{1}\cdots u_{8}]\subset
R\mathbb{\mathbb{T}}_{8}$.

The standard maximal torus $\rm T^{8}\subset\mathrm{E}_{8}$ is the image of
$\widetilde{\rm T}_{8}\subset\mathrm{Spin}(16)$ under the homomorphism
$\mathrm{Spin}(16)\to\mathrm{E}_{8}$. Therefore $\rm T^{8}=\widetilde{\rm
T}_{8}/\langle\zeta\rangle$ where $\zeta\in\mathrm{Spin}(16)$ is the central
element $e_{1}\cdots e_{16}\in\mathrm{Spin}(16)$. Thus $R{\rm T}^{8}\subset
R\widetilde{\rm T}_{8}$ is generated by those characters of $\widetilde{\rm
T}_{8}$ which map $\zeta$ to $1$. Since $u_{j}^{2}(\zeta)=-1\in\mathbb{S}^{1}$
for each $j$, we see that $u_{i}^{2}u_{j}^{\pm 2},u_{i}^{-2}u_{j}^{\pm 2}\in
R\rm T^{8},1\leq i<j\leq 8$. Also $u_{1}^{\varepsilon_{1}}\cdots
u_{8}^{\varepsilon_{8}}$ are in $R\rm T^{8}$ when $\sum_{1\leq j\leq
8}\varepsilon_{j}$ is even, with $\varepsilon_{j}\in\\{1,-1\\}$. In fact,
these 240 characters generate the ring $R\rm T^{8}$. To see this, Note that
these are precisely the non-zero weights of the (complexified) adjoint
representation $\mathfrak{e}_{8}:=Lie(\mathrm{E}_{8})\otimes\mathbb{C}$ of
$\mathrm{E}_{8}$ since
$\mathfrak{e}_{8}=(Lie(\mathrm{Spin}(16))\otimes\mathbb{C})\oplus\Delta_{8}^{+}$
as a representation of $\mathrm{Spin}(16)$. Since $\mathrm{E}_{8}$ is simply
connected and has trivial centre, we have $P_{8}=Q_{8}$ where $P_{8},Q_{8}$
are the weight lattice and the root lattice of $\mathrm{E}_{8}$, respectively.
As for any connected compact semisimple Lie group $G$ we have
$\mathbb{Z}Q_{8}\subset R\rm T^{8}\subset\mathbb{Z}P_{8}$ and so we must have
equality throughout. (Here, for group $A$, $\mathbb{Z}A$ denotes the group
ring of $A$.) This shows that $R\rm T^{8}=\mathbb{Z}Q_{8}$ as asserted.

Next, we describe $R\rm T^{6}$ as a quotient of $R\rm T^{8}$. We have the
commutative diagram where the copy of $\mathbb{S}^{1}$ in ${\rm
T}^{8}\subset\mathrm{E}_{8}$ is described in 2.1.

${\widetilde{\rm T}_{8}}$${\rm T^{8}}$${\mathrm{E}_{8}}$${\widetilde{\rm
T}_{5}\times\mathbb{S}^{1}}$${\rm T^{6}}$${\mathrm{E}_{6}}$ Figure 3. Maximal
tori of $\mathrm{E}_{6}$ and $\mathrm{E}_{8}$.

Let $\xi$ be the standard (one dimensional) representation of
$\mathbb{S}^{1}$, corresponding to the identity map of $\mathbb{S}^{1}$. Thus
$R\mathbb{S}^{1}=\mathbb{Z}[\xi,\xi^{-1}].$

The discussion from §2.1 shows that the restriction map $R\widetilde{\rm
T}_{8}=\mathbb{Z}[u_{1}^{\pm 2},\cdots,u_{8}^{\pm 2},u_{1}\cdots
u_{8}]\to\mathbb{Z}[\xi,\xi^{-1}]=R\mathbb{S}^{1}$ is given by $u_{j}\mapsto
1$ for $1\leq j\leq 5$ and $u_{j}\mapsto\xi,6\leq j\leq 8$. So the restriction
map $R\rm T^{8}\to R\mathbb{S}^{1}$ is essentially given by the same formula.
Now the homomorphism $R\rm T^{8}\to R(\widetilde{\rm
T}_{5}\times\mathbb{S}^{1})=R\widetilde{\rm T}_{5}\otimes
R\mathbb{S}^{1}=\mathbb{Z}[u_{1}^{\pm 2},\ldots,u_{5}^{\pm 2},u_{1}\cdots
u_{5},\xi,\xi^{-1}]$ is defined by

(2) $u_{i}\mapsto u_{i},1\leq i\leq 5,\textrm{\leavevmode\nobreak\
and\leavevmode\nobreak\ }u_{j}\mapsto\xi,6\leq j\leq 8.$

In particular, $\prod_{1\leq j\leq 8}u_{j}$ restricts to $u_{1}\cdots
u_{5}\xi^{3}.$ Since $\widetilde{\rm T}_{5}\times\mathbb{S}^{1}\to\rm T^{6}$
is a covering projection, $R{\rm T}^{6}\to R(\widetilde{\rm
T}_{5}\times\mathbb{S}^{1})$ is a monomorphism, which is regarded as an
inclusion map. The ring $R{\rm T}^{6}\subset R(\widetilde{\rm
T}_{5}\times\mathbb{S}^{1})$ is generated by the characters which are trivial
on $(e_{1}\cdots e_{10},i)\in\widetilde{\rm T}_{5}\times\mathbb{S}^{1}.$ Since
$\widetilde{\rm T}_{5}\times\mathbb{S}^{1}\to\rm T^{8}$ factors through $\rm
T^{6}\hookrightarrow\rm T^{8}$, it follows that the restriction homomorphism
$R{\rm T}^{8}\to R\rm T^{6}$ is given by the same formula (2) as above. In
particular, $\Delta^{+}_{8}\in R\rm T^{8}$ restricts to
$\Delta_{5}^{+}\xi^{3}+\Delta^{-}_{5}\xi^{-3}+3\Delta_{5}^{+}\xi^{-1}+3\Delta_{5}^{-}\xi\in
R\rm T^{6}$.

###  3.1. Roots, weights, and fundamental representations of $\mathrm{E}_{6}$

Since $\mathrm{E}_{6}$ is simply connected, the representation ring
$R\mathrm{E}_{6}$ is a polynomial algebra in the classes of the fundamental
representations of $\mathrm{E}_{6}$. We shall use a different set of
polynomial generators of $R\mathrm{E}_{6}$ which are more convenient for
describing the restriction homomorphism $R\mathrm{E}_{6}\to
R\mathrm{Spin}(10)$. See Proposition 3.1.

We begin by describing the roots and the fundamental weights of
$\mathrm{E}_{6}$. For a more detailed discussion, we refer the reader to [3].

Denote by $x_{j}/2:Lie(\mathbb{T}_{8})\to\mathbb{R}$ the linear map such that
$u_{j}(t)=\exp(\pi ix_{j}(t))\in\mathbb{S}^{1}$ for all $t\in
Lie(\mathbb{T}_{8})=\mathbb{R}^{8}$. Then, the roots of $E_{8}$ are obtained
as $x_{i}\pm x_{j},-(x_{i}\pm x_{j}),1\leq i<j\leq 8$, $(\sum_{1\leq j\leq
8}\varepsilon_{j}x_{j})/2$, where $\varepsilon_{j}\in\\{1,-1\\}$ with
$\sum\varepsilon_{j}$ even.

The root lattice of $\mathrm{SU}(3)\subset\mathrm{E}_{8}$ is the subgroup
generated by $x_{6}-x_{7},x_{7}-x_{8}\in Q_{8}$. Since $\mathrm{E}_{6}$ is the
centralizer of $\mathrm{SU}(3)$, the root lattice $Q_{6}$ of $\mathrm{E}_{6}$
is obtained as the subgroup of vectors in $Q_{8}\subset\mathbb{R}^{8}$ that
are orthogonal, with respect to the standard inner product on
$\mathbb{R}^{8}$, to $\\{x_{6}-x_{7},x_{7}-x_{8}\\}$. Thus the set $\Phi_{6}$
of roots of $\mathrm{E}_{6}$ consists of $\pm(x_{i}\pm x_{j}),1\leq i<j\leq
5,(\sum_{1\leq j\leq 8}\varepsilon_{j}x_{j})/2$ where $\sum\varepsilon_{j}$ is
even, $\varepsilon_{j}\in\\{1,-1\\}$ with
$\varepsilon_{6}=\varepsilon_{7}=\varepsilon_{8}.$ We set
$\alpha_{1}=\frac{1}{2}(x_{1}+x_{2}-x_{3}-x_{4}-x_{5}-x_{6}-x_{7}-x_{8}),\alpha_{2}=-x_{2}+x_{3},\alpha_{3}=x_{2}+x_{3},\alpha_{4}=-x_{3}+x_{4},\alpha_{5}=-x_{4}+x_{5},\alpha_{6}=\frac{1}{2}(x_{1}-x_{2}-x_{3}-x_{4}-x_{5}+x_{6}+x_{7}+x_{8})$.
Then $\\{\alpha_{i}\mid 1\leq i\leq 6\\}$ is a set of simple roots where the
labelling is consistent with the labelling of the nodes of the Dynkin diagram
of $\mathrm{E}_{8}$. We shall denote the set of positive roots of
$\mathrm{E}_{6}$ by $\Phi_{6}^{+}.$ The corresponding fundamental weights of
$\mathrm{E}_{6}$ are

$\displaystyle\varpi_{1}$ $\displaystyle=x_{1}-\frac{1}{3}(x_{6}+x_{7}+x_{8})$
$\displaystyle\varpi_{2}$
$\displaystyle=\frac{3}{2}x_{1}+\frac{1}{2}(-x_{2}+x_{3}+x_{4}+x_{5})-\frac{1}{6}(x_{6}+x_{7}+x_{8})$
$\displaystyle\varpi_{3}$
$\displaystyle=\frac{3}{2}x_{1}+\frac{1}{2}(x_{2}+x_{3}+x_{4}+x_{5})+\frac{1}{6}(x_{6}+x_{7}+x_{8})$
$\displaystyle\varpi_{4}$ $\displaystyle=2x_{1}+x_{4}+x_{5}$
$\displaystyle\varpi_{5}$ $\displaystyle=x_{1}+x_{5}$
$\displaystyle\varpi_{6}$
$\displaystyle=x_{1}+\frac{1}{3}(x_{6}+x_{7}+x_{8}).$

Therefore the weight lattice $P_{6}$ of $\mathrm{E}_{6}$ equals
$\\{\sum_{1\leq j\leq 6}a_{j}\varpi_{j}\mid a_{j}\in\mathbb{Z}\\}$. Observe
that, since $\varpi_{5}=x_{1}+x_{5}$ is a root, $V_{\varpi_{5}}$ is the
(complexified) adjoint representation $\mathfrak{e}_{6}.$ The dimensions of
the fundamental representations of $\mathrm{E}_{6}$ are well-known to experts
and are given in Table 1. (Cf. http://oeis.org/A121737.) In any case, the
dimension of an irreducible representation $V_{\lambda}$ with highest weight
$\lambda$ can be obtained using the Weyl dimension formula. See [11, Theorem
10.18].

The involution of the Dynkin diagram of $\mathrm{E}_{6}$ fixes the nodes $4,5$
and swaps $1$ with $6$ and $2$ with $3$. The outer automorphism of
$\mathrm{E}_{6}$, that induces the diagram automorphism, induces an involution
of $R\mathrm{E}_{6}$ that fixes $[V_{\varpi_{4}}],[V_{\varpi_{5}}]$ and swaps
$[V_{\varpi_{1}}]$ with $[V_{\varpi_{6}}]$ and $[V_{\varpi_{2}}]$ with
$[V_{\varpi_{3}}]$.

Highest weight |  Representation |  Dimension
---|---|---
$\varpi_{1},\varpi_{6}$ |  $V_{\varpi_{1}},V_{\varpi_{6}}$ |  27
$\varpi_{2},\varpi_{3}$ |  $V_{\varpi_{2}},V_{\varpi_{3}}$ |  351
$\varpi_{5}$ |  $V_{\varpi_{5}}$ |  78
$\varpi_{4}$ |  $V_{\varpi_{4}}$ |  2925

Table 1. Dimensions of fundamental representations of $\mathrm{E}_{6}$.

As mentioned already, since $\mathrm{E}_{6}$ is simply connected, we have
$R\mathrm{E}_{6}=\mathbb{Z}[[V_{\varpi_{1}}],[V_{\varpi_{2}}],\dots,[V_{\varpi_{6}}]]$,
where $V_{\varpi_{j}}$ is complex representation of $\mathrm{E}_{6}$ with
highest weight $\varpi_{j},1\leq j\leq 6$. Proposition 3.1 below says that the
generators $[V_{\varpi_{4}}]$ of $R\mathrm{E}_{6}$ can be replaced by
$\Lambda^{2}(V_{\varpi_{5}})$. This is a particular case of a more general
result of Adams [2, Proposition 1], detailed proof of which can be found in
the paper of Guillot [10, §4].

######  Proposition 3.1.

Set
$\alpha:=[V_{\varpi_{1}}],\beta:=[V_{\varpi_{6}}],\gamma:=[V_{\varpi_{5}}]$.
Then $[V_{\varpi_{2}}]=\Lambda^{2}(\alpha)$,
$[V_{\varpi_{3}}]=\Lambda^{2}(\beta)$, and
$[V_{\varpi_{4}}]=\Lambda^{2}(\gamma)-\gamma$ and
$\Lambda^{3}(V_{\varpi_{1}})\cong
V_{\varpi_{4}}\cong\Lambda^{3}(V_{\varpi_{6}}).$ In particular,

$R\mathrm{E}_{6}=\mathbb{Z}[\alpha,\beta,\gamma,\Lambda^{2}(\alpha),\Lambda^{2}(\beta),\Lambda^{2}(\gamma)].$

###### Proof.

It is easily seen that the highest weight that occurs in the representation
$\Lambda^{2}(V_{\varpi_{1}})$ is
$\varpi_{1}+\varpi_{1}-\alpha_{1}=2\varpi_{1}-\alpha_{1}=\varpi_{2}$. It
follows that the irreducible representation $V_{\varpi_{2}}$ occurs as a
subrepresentation of $\Lambda^{2}(V_{\varpi_{1}})$. From Table 1, we get that
$\dim(V_{\varpi_{2}})=351=\binom{27}{2}=\dim\Lambda^{2}(V_{\varpi_{1}})$.
Hence $\Lambda^{2}(V_{\varpi_{1}})=V_{\varpi_{2}}$. The same argument shows
that $\Lambda^{2}(V_{\varpi_{6}})=V_{\varpi_{3}}$. As remarked already,
$V_{\varpi_{5}}=\mathfrak{e}_{6}$ is the (complexified) adjoint representation
$\mathrm{E}_{6}$. It follows that $\Lambda^{2}(V_{\varpi_{5}})$ contains
$V_{\varpi_{5}}$ as a subrepresentation. In fact, the Lie bracket defines an
alternating bilinear map
$\mathfrak{e}_{6}\mathfrak{\times}\mathfrak{e}_{6}\to\mathfrak{e}_{6}$ namely
$(X,Y)\mapsto[X,Y]$. Thus we obtain homomorphism
$\Lambda^{2}(V_{\varpi_{5}})\to V_{\varpi_{5}}$ which is a morphism of
$\mathrm{E}_{6}$-modules. This is a surjection since $\mathfrak{e}_{6}$ is a
simple Lie algebra. By complete reducibility we obtain that $V_{\varpi_{5}}$
is a subrepresentation of $\Lambda^{2}(V_{\varpi_{5}})$. Again, weight
consideration shows that
$\varpi_{5}+(\varpi_{5}-\alpha_{5})=2\varpi_{5}-\alpha_{5}=\varpi_{4}$ is the
highest weight that occurs in $\Lambda^{2}(V_{\varpi_{5}})$ and so
$V_{\varpi_{4}}$ is a subrepresentation of $\Lambda^{2}(V_{\varpi_{5}})$.
Furthermore, $\dim(\Lambda^{2}(V_{\varpi_{5}}))={78\choose
2}=3003,\dim(V_{\varpi_{5}})=78$ and $\dim(V_{\varpi_{4}})$ is known to be
$2925$. It follows that

(3) $\Lambda^{2}(V_{\varpi_{5}})\cong V_{\varpi_{4}}\oplus V_{\varpi_{5}}.$

We conclude that
$R\mathrm{E}_{6}=\mathbb{Z}[\alpha,\beta,\gamma,\Lambda^{2}(\alpha),\Lambda^{2}(\beta),\Lambda^{2}(\gamma)]$.

It remains to show that
$\Lambda^{3}(V_{\varpi_{1}})\cong\Lambda^{3}(V_{\varpi_{6}})\cong
V_{\varpi_{4}}.$ We know that $\dim
V_{\varpi_{4}}=\dim\Lambda^{3}(V_{\varpi_{1}})=\dim\Lambda^{3}(V_{\varpi_{6}})=2925$
and that $3\varpi_{1}-2\alpha_{1}-\alpha_{2}$ (resp.
$3\varpi_{6}-2\alpha_{6}-\alpha_{1})$ is a (highest) weight of
$\Lambda^{3}(V_{\varpi_{1}})$ (resp. $\Lambda^{3}(V_{\varpi_{6}}))$. Since
$3\varpi_{1}-2\alpha_{1}-\alpha_{2}=\varpi_{4}=3\varpi_{6}-2\alpha_{6}-\alpha_{3}$,
arguing as before, we have
$V_{\varpi_{4}}\cong\Lambda^{3}(V_{\varpi_{1}})\cong\Lambda^{3}(V_{\varpi_{6}})$.
Equation (3) can be rewritten as

(4) $\Lambda^{2}(V_{\varpi_{5}})\cong\Lambda^{3}(V_{\varpi_{6}})\oplus
V_{\varpi_{5}}.$

∎

We end this section with the computation of the restriction of the
(complexified) adjoint representation
$\mathfrak{e}_{8}=Lie(\mathrm{E}_{8})\otimes\mathbb{C}$ of $\mathrm{E}_{8}$ to
$\mathrm{E}_{6}.$ This will aid in describing certain fundamental
representations of $\mathfrak{e}_{6}$ to
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}.$ Although the result is known—see
[3]—we give the details for the sake of completeness.

Since the maximal torus $\rm T^{6}$ is contained in
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}\subset\mathrm{E}_{6}$, we need only
describe the restriction of $\mathfrak{e}_{8}$ to
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$. Since
$Lie(\mathrm{E}_{8})=Lie(\mathrm{Spin}(16))\oplus\Delta_{8,\mathbb{R}}^{+}$,
this amounts to finding the restrictions of the
$\mathrm{Spin}(16)$-representation
$Lie(\mathrm{Spin}(16))\otimes\mathbb{C}\oplus\Delta_{8}^{+}$ to
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}\subset\mathrm{Spin}(16)$.

Working in $R\mathbb{T}_{8}$, the characters $u_{j},j=6,7,8$ restrict to the
identity character $\xi$ of $\mathbb{S}^{1}$ and $u_{i},1\leq i\leq 5,$ to the
trivial character on $\mathbb{S}^{1}$. Each $u_{j},j=6,7,8,$ restricts to
$\xi$ on the $\mathbb{S}^{1}$ factor. Thus
$u_{6}^{\varepsilon_{6}}u_{7}^{\varepsilon_{7}}u_{8}^{\varepsilon_{8}}$
restricts to $\xi^{\varepsilon_{6}+\varepsilon_{7}+\varepsilon_{8}}$. Also
$u_{i}^{2}u_{j}^{\pm 2}\mapsto u_{i}^{2}\xi^{\pm 2}$ for $1\leq i\leq 5<j\leq
8$, and, $u_{i}^{2}u_{j}^{-2}\mapsto 1,u_{i}^{2}u_{j}^{2}\mapsto\xi^{\pm 4}$
for $6\leq i<j\leq 8$. It follows that $\Delta^{+}_{8}$ restricts to
$\Delta_{5}^{+}\xi^{3}\oplus\Delta_{5}^{-}\xi^{-3}\oplus
3\Delta_{5}^{+}\xi^{-1}\oplus 3\Delta_{5}^{-}\xi^{1}$ on
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$.

Among the roots of $Lie(\mathrm{Spin}(16))$, $(\pm x_{i}\pm x_{j}),1\leq
i<j\leq 5,$ restrict to the roots of $Lie(\mathrm{Spin}(10))$. For each fixed
$j,6\leq j\leq 8$, the set of roots $(\pm x_{i}+x_{j}),1\leq i\leq 5,$
correspond to the characters $u_{i}^{\pm 2}\xi^{2}$ which are precisely the
characters of $\lambda_{1}\otimes\mathbb{C}_{\xi^{2}}$, where
$\lambda_{1}\cong\mathbb{C}^{10}\mathbb{=}\mathbb{R}^{10}\otimes\mathbb{C}$ is
the standard 10-dimensional representation on which $\mathrm{Spin}(10)$ acts
via the double covering $\mathrm{Spin}(10)\to\mathrm{SO}(10)$. Similarly the
roots $\pm x_{i}-x_{j},1\leq i\leq 5<j\leq 8$, correspond to the characters of
$\mathbb{C}^{10}\otimes\mathbb{C}_{\xi^{-2}}.$ Of the remaining roots, each of
the six roots $\pm(x_{6}-x_{7}),\pm(x_{6}-x_{8}),\pm(x_{7}-x_{8})$ yields the
trivial representations, and each of
$\varepsilon(x_{6}+x_{7}),\varepsilon(x_{6}+x_{8}),\varepsilon(x_{7}+x_{8})$
yields the representation $\xi^{4\varepsilon},\varepsilon\in\\{1,-1\\}$. The
trivial character occurs in $Lie(\mathrm{Spin}(16))$ with multiplicity
$8=\mathrm{rank}(\mathrm{Spin}(16))$ whereas it occurs in
$Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})$ with multiplicity $6$. All told,
we obtained that $Lie(\mathrm{Spin}(16))\otimes\mathbb{C}$ restricts to
$Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\otimes\mathbb{C}\oplus
3\lambda_{1}\xi^{2}\oplus 3\lambda_{1}\xi^{-2}\oplus 3\xi^{4}\oplus
3\xi^{-4}\oplus 8$.

In the sequel, we shall denote by $\lambda_{j},1\leq j\leq 5$ the
representation $\Lambda^{j}(\lambda_{1}).$

From the above discussion, we conclude that, denoting by $\rho^{\prime}$ the
restriction $R\mathrm{E}_{8}\to R(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})$,

(5)
$\rho(\mathfrak{e}_{8})=Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\otimes\mathbb{C}\oplus\Delta_{5}^{+}\xi^{3}\oplus\Delta_{5}^{-}\xi^{-3}\oplus\mathbb{C}^{8}\oplus
3(V\oplus V^{\prime})$

where $V=\lambda_{1}\xi^{-2}\oplus\Delta_{5}^{-}\xi\oplus\xi^{4},$ and
$V^{\prime}=\lambda_{1}\xi^{2}\oplus\Delta_{5}^{+}\xi^{-1}\oplus\xi^{-4}$ are
representations of $\mathrm{Spin}(10)\cdot\mathbb{S}^{1}.$

Using the fact that the coefficients of $x_{j},6\leq j\leq 8,$ in the roots of
$\mathfrak{e}_{6}$ are all equal, it is easily verified that

(6) $\mathfrak{e}_{6}\cong
Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\otimes\mathbb{C}\oplus\Delta_{5}^{+}\xi^{3}\oplus\Delta_{5}^{-}\xi^{-3}.$

It follows from (5) that $V\oplus V^{\prime}$ are representations of
$\mathrm{E}_{6}$.

Claim:

(7) $V_{\varpi_{1}}\cong
V=\lambda_{1}\xi^{-2}\oplus\Delta_{5}^{-}\xi\oplus\xi^{4};\leavevmode\nobreak\
V_{\varpi_{6}}\cong
V^{\prime}=\lambda_{1}\xi^{2}\oplus\Delta_{5}^{+}\xi^{-1}\oplus\xi^{-4}.$

It is readily seen that (i) $\dim V=27,$ and, (ii)
$x_{1}-(1/3)(x_{6}+x_{7}+x_{8})$ is a weight of $V$— in fact it is the highest
weight of $V.$ It turns out that $V$ is in fact a representation of
$\mathrm{E}_{6}$ isomorphic to $V_{\varpi_{1}}$. Similarly,
$V^{\prime}=\lambda_{1}\xi^{2}\oplus\Delta_{5}^{+}\xi^{-1}\oplus\xi^{-4}$ is
isomorphic to $V_{\varpi_{6}}$. Thus Equation (5) can be rewritten as

(8) $\tilde{\rho}(\mathfrak{e}_{8})=\mathfrak{e}_{6}\oplus
3(V_{\varpi_{1}}\oplus V_{\varpi_{6}})\oplus\mathbb{C}^{8}$

where $\tilde{\rho}:R\mathrm{E}_{8}\to R\mathrm{E}_{6}$ is the restriction
homomorphism.

##  4\. The Restriction Homomorphism $\rho:R\mathrm{E}_{6}\to
R\mathrm{Spin}(10)$

We keep the notations of the previous section. Set
$H:=\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$. We shall describe the restriction
homomorphism $\rho^{\prime}:R\mathrm{E}_{6}\to RH$. Although our interest here
is in the restriction $\rho:R\mathrm{E}_{6}\to R\mathrm{Spin}(10),$ the
computations of $\rho^{\prime}$ will be more transparent since both
$\mathrm{E}_{6}$ and $H$ have the same rank. We have ([14, §12])
$R\mathrm{Spin}(10)=\mathbb{Z}[\lambda_{1},\lambda_{2},\lambda_{3},\Delta^{+}_{5},\Delta^{-}_{5}]$.
We denote by $\pi_{j},$ the highest weights of $\lambda_{j},1\leq j\leq 4.$
Then $\pi_{1},\pi_{2},$ and $\pi_{3}$ are fundamental weights. Note that
$V_{\pi_{1}}$ is the representation of $\mathrm{Spin}(10)$ obtained via
$\pi:\mathrm{Spin}(10)\to\mathrm{SO}(10)$ from the standard representation of
$\mathrm{SO}(10)$ on $U:=\mathbb{C}^{10}$.

Although $V_{\pi_{4}}=\Lambda^{4}(U)$ is irreducible, it is not a fundamental
representation.

$RH$ is the subring of
$R(\mathrm{Spin}(10)\times\mathbb{S}^{1})=\mathbb{Z}[\lambda_{1},\lambda_{2},\lambda_{3},\Delta_{5}^{\pm},\xi^{\pm}]$
generated by the elements $[V]$ where $V$ is a representation of
$\mathrm{Spin}(10)\times\mathbb{S}^{1}$ on which $(e_{1}\cdots e_{10},i)$ acts
trivially. Of course, any representation of $\mathrm{E}_{6}$ when restricted
to $H$ satisfies this requirement.

Recall from (7) that $V=V_{\varpi_{1}}$ and $V^{\prime}=V_{\varpi_{6}}$
restrict to the representations
$V=\lambda_{1}\xi^{-2}+\Delta_{5}^{-}\xi+\xi^{4}$ and
$V^{\prime}=\lambda_{1}\xi^{2}+\Delta_{5}^{+}\xi^{-1}+\xi^{-4}$ on
$\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$. Since $\mathfrak{e}_{6}\cong
Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\otimes\mathbb{C}\oplus\Delta_{5}^{+}\xi^{3}\oplus\Delta_{5}^{-}\xi^{-3}$,
as a representation of $\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$, and since
$1+\lambda_{2}=Lie(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\otimes\mathbb{C},$ we
have

(9)
$\rho^{\prime}(V_{\varpi_{5}})=1+\lambda_{2}+\Delta^{+}_{5}\xi^{3}+\Delta^{-}_{5}\xi^{-3}.$

The restriction map $\rho^{\prime}\colon R\mathrm{E}_{6}\to
R(\mathrm{Spin}(10)\cdot S^{1})$ is a $\lambda$-ring homomorphism and
observing that
$\Lambda^{2}(\Delta_{5}^{+})=\lambda_{3}=\Lambda^{2}(\Delta_{5}^{-})$, routine
calculation shows that

(10)
$\rho^{\prime}(\Lambda^{2}(V_{\varpi_{1}}))=\lambda_{2}\xi^{-4}+\lambda_{3}\xi^{2}+\lambda_{1}\Delta_{5}^{-}\xi^{-1}+\lambda_{1}\xi^{2}+\Delta_{5}^{-}\xi^{5},$
(11)
$\rho^{\prime}(\Lambda^{2}(V_{\varpi_{6}}))=\lambda_{2}\xi^{4}+\lambda_{3}\xi^{-2}+\lambda_{1}\Delta_{5}^{+}\xi+\lambda_{1}\xi^{-2}+\Delta_{5}^{+}\xi^{-5},$
(12)
$\rho^{\prime}(\Lambda^{2}(V_{\varpi_{5}}))=\lambda_{2}+(1+\lambda_{2})(\Delta_{5}^{+}\xi^{3}+\Delta_{5}^{-}\xi^{-3})+\lambda_{3}\xi^{6}+\lambda_{3}\xi^{-6}+\Lambda^{2}(\lambda_{2})+\Delta_{5}^{+}\Delta_{5}^{-}.$

The following lemma allows us to express the term $\Lambda^{2}(\lambda_{2})$
in the last equation in terms of $\lambda_{1}\lambda_{3}$ and $\lambda_{4}$.

######  Lemma 4.1.

The relation

(13) $\lambda_{1}\lambda_{3}=\Lambda^{2}(\lambda_{2})+\lambda_{4}$

holds in the ring $R\mathrm{Spin}(10)$.

###### Proof.

Let $U_{j}:=\Lambda^{j}(\mathbb{C}^{10}),$ the representation of
$\mathrm{Spin}(10)$ with highest weight $\pi_{j},$ $1\leq j\leq 4.$

We have an obvious surjective homomorphism $f:U_{1}\otimes U_{3}\to U_{4}$ of
$\mathrm{Spin}(10)$-representations defined as $u\otimes v\mapsto u\wedge v$.
In fact the vector $w_{1}=e_{1}\otimes e_{2}e_{3}e_{4}-e_{2}\otimes
e_{1}e_{3}e_{4}+e_{3}\otimes e_{1}e_{2}e_{4}-e_{4}\otimes e_{1}e_{2}e_{3}$ is
seen to be a highest weight vector of $U_{1}\otimes U_{3}$ that maps to
$4e_{1}e_{2}e_{3}e_{4}.$ Here, and in what follows, we have written $uv$ to
denote $u\wedge v.$ On the other hand $w_{0}:=e_{1}\otimes e_{1}e_{2}e_{3}\in
U_{1}\otimes U_{3}$ is a highest weight vector whose weight equals
$\pi_{1}+\pi_{3}=\pi_{1}+3\pi_{1}-2\alpha_{1}-\alpha_{2}-\alpha_{3}=(2\pi_{1}-\alpha_{1})+(2\pi_{1}-\alpha_{1}-\alpha_{2})=:\omega$
which is the highest among all the weights of $\Lambda^{2}(U_{2}).$ Since
$w_{0}\in\ker(f)$ and $f$ is a homomorphism of
$\mathrm{Spin}(10)$-representation, the $\mathrm{Spin}(10)$-submodule
generated by $w_{0}$, denoted $W$ is contained in $U_{1}\otimes U_{3}$ and is
linearly disjoint with $U_{4}$. So $W\oplus U_{4}$ is a subrepresentation of
$U_{1}\otimes U_{3}$. It turns out that $\dim W=990=\dim\Lambda^{2}(U_{2}).$
This may be verified by applying the Weyl dimension formula. It follows that
$\dim W+\dim U_{4}=990+210=1200=\dim(U_{1}\otimes U_{3}).$ So we must have
$W\oplus U_{4}\cong U_{1}\otimes U_{3}.$ This completes the proof of the
lemma. ∎

Another approach to proving the above lemma which is conceptually simpler but
computationally more involved, is to show the equality of the characters on
both sides of (13). This may be done by directly calculating the case at hand
rather than using the Weyl character formula. Starting with the standard basis
for $U_{1}=\mathbb{C}^{10},$ has natural bases of $U_{2},U_{3},U_{4}$ and also
the tensor products $U_{1}\otimes U_{3}$ as well as $\Lambda^{2}(U_{2})$,
consisting of weight vectors. It is straightforward to count with multiplicity
the weights of these weight vectors. The calculations are tabulated below, but
the details are left to the reader. Indeed, this was the proof given in [16].

Weight type |  No. of weights |  Mult. in $\Lambda^{2}(U_{2})$ |  Mult. in $U_{4}$ |  Mult. in $U_{1}\otimes U_{3}$
---|---|---|---|---
$\pm x_{i}\pm x_{j}\pm x_{k}\pm x_{l};$ |  $80$ |  $3$ |  $1$ |  $4$
$i<j<k<l$ |  |  |  |
$\pm 2x_{i}\pm x_{j}\pm x_{k};$ |  $240$ |  $1$ |  $0$ |  $1$
$i<j<k.$ |  |  |  |
$\pm x_{i}\pm x_{j};i<j.$ |  $40$ |  $11$ |  $3$ |  $14$
$\pm 2x_{i}$ |  $10$ |  $4$ |  $0$ |  $4$
$0$ |  $1$ |  $30$ |  $10$ |  $40$

Table 2. Weights of $\Lambda_{2}(U_{2})$ and $U_{4}$.

Recall that the relation
$\Delta_{5}^{+}\Delta_{5}^{-}=1+\lambda_{2}+\lambda_{4}$ holds in
$\mathrm{Spin}(10)$; see [14, Theorem 10.3, Chapter 13]. Therefore, from the
above lemma and Equation (12), we get that

(14)
$\rho^{\prime}(\Lambda^{2}(\gamma))=1+2\lambda_{2}+(1+\lambda_{2})(\Delta_{5}^{+}\xi^{3}+\Delta_{5}^{-}\xi^{-3})+\lambda_{3}\xi^{6}+\lambda_{3}\xi^{-6}+\lambda_{1}\lambda_{3}.$

Let $i^{*}:R(\mathrm{Spin}(10)\cdot\mathbb{S}^{1})\to R\mathrm{Spin}(10)$
denote the homomorphism of rings induced by the inclusion
$i\colon\mathrm{Spin}(10)\to\mathrm{Spin}(10)\cdot\mathbb{S}^{1}$. Then
$\rho=i^{*}\circ\rho^{\prime}\colon R\mathrm{E}_{6}\to R\mathrm{Spin}(10)$ is
the restriction homomorphism induced by the inclusion
$\mathrm{Spin}(10)\to\mathrm{E}_{6}$. Then $\rho$ is obtained from Equations
(7), (9), (10), (11), and (14), using the fact that $i^{*}(\xi)=1$, the
trivial 1-dimensional representation of $R\mathrm{Spin}(10)$. Therefore, we
obtain the following.

######  Proposition 4.2.

The restriction homomorphism $\rho:R\mathrm{E}_{6}\to R\mathrm{Spin}(10)$ is
given as follows. (i)
$\alpha^{\prime}\colon=\rho(\alpha)=1+\lambda_{1}+\Delta_{5}^{+},$
(ii) $\beta^{\prime}\colon=\rho(\beta)=1+\lambda_{1}+\Delta_{5}^{-},$
(iii)
$\gamma^{\prime}\colon=\rho(\gamma)=1+\lambda_{2}+\Delta_{5}^{+}+\Delta_{5}^{-},$
(iv)
$\rho(\Lambda^{2}(\alpha))=\lambda_{2}+\lambda_{3}+\lambda_{1}\Delta_{5}^{-}+\lambda_{1}+\Delta_{5}^{-},$
(v)
$\rho(\Lambda^{2}(\beta))=\lambda_{2}+\lambda_{3}+\lambda_{1}\Delta_{5}^{+}+\lambda_{1}+\Delta_{5}^{+},$
(vi)
$\rho(\Lambda^{2}(\gamma))=1+2\lambda_{2}+(1+\lambda_{2})(\Delta_{5}^{+}+\Delta_{5}^{-})+2\lambda_{3}+\lambda_{1}\lambda_{3}$.
$\blacksquare$

##  5\. Proofs of Theorems 1.1 and 1.2

We begin with a brief description of the Hodgkin spectral sequence and the
change of rings spectral sequence.

###  5.1. Hodgkin Spectral sequence

We shall now describe briefly the Hodgkin spectral sequence. For a more
detailed exposition we refer the reader to [18], [4], [6], or [17].

Let $G$ be a compact connected Lie group and $H$ be a closed subgroup. Let
$\rho:RG\to RH$ be the restriction homomorphism and let $RG\to\mathbb{Z}$ be
the augmentation. We regard $RH$ and $\mathbb{Z}$ as $RG$-modules via these
homomorphisms. The following theorem is due to Hodgkin [13].

######  Theorem 5.1.

Let $G$ be a compact Lie group and $H$ an closed subgroup of $G$. Suppose that
$\pi_{1}(G)$ is torsion-free. There exists a spectral sequence that converges
to $K^{*}(G/H)$ whose $E_{2}$-page is

$E_{2}^{p,q}=\begin{cases}\mathrm{Tor}^{p}_{RG}(RH,\mathbb{Z}),&\textrm{
\leavevmode\nobreak\ if\leavevmode\nobreak\ }q\equiv 0\pmod{2}\\\
0,&\textrm{\leavevmode\nobreak\ otherwise.}\end{cases}$

The differential $d_{r}$ has bidegree $(r,1-r)$. In particular, $d_{r}=0$ for
all $r$ even.

In the above theorem, $\mathrm{Tor}^{p}_{R}(M,N)$ is, by definition,
$\mathrm{Tor}_{-p}^{R}(M,N)$. In order to compute
$\mathrm{Tor}^{*}_{RG}(RH,\mathbb{Z})$, one uses the ‘change of rings’
theorem. This is a spectral sequence, referred to as the change of ring
spectral sequence in the sequel. Its $E_{2}$-page is defined as follows:
$E_{2}^{p,q}=\mathrm{Tor}_{\Omega}^{p}(\mathrm{Tor}_{\Lambda}^{q}(RH,\mathbb{Z}),\mathbb{Z})$
where $\Lambda$ is a conveniently chosen subring of $RG$ and $\Omega$ is the
quotient of $RG$ by the ideal generated by the kernel of the augmentation
$\Lambda\to\mathbb{Z}$. The spectral sequence converges to
$\mathrm{Tor}^{*}_{RG}(RH,\mathbb{Z})$. This is a special case of a very
general result established in [7, p. 379].

We shall find a subalgebra $\Lambda$ of $R\mathrm{E}_{6}$ over which
$R\mathrm{Spin}(10)$ is free. This will allow us to apply the change of ring
theorem to simplify the computation of
$\mathrm{\mathrm{Tor}}^{*}_{R\mathrm{E}_{6}}(R\mathrm{Spin}(10),\mathbb{Z}).$
The appropriate choice of $\Lambda$ is given in the following lemma.

######  Lemma 5.2.

Let $\Lambda$ be the subring
$\mathbb{Z}[\alpha,\beta,\gamma,\Lambda^{2}(\alpha)]$ of $R\mathrm{E}_{6}$.
Then $R\mathrm{Spin}(10)$ is a free $\Lambda$-module (via the restriction
homomorphism) with basis $\\{\lambda_{1}^{i}:i\geq 0\\}$ over $\Lambda$.

###### Proof.

Let
$\Lambda^{\prime}\colon=\rho(\Lambda)=\mathbb{Z}[\alpha^{\prime},\beta^{\prime},\gamma^{\prime},\Lambda^{2}(\alpha^{\prime})]$.
Let us write
$\gamma^{\prime\prime}=\gamma^{\prime}-\alpha^{\prime}-\beta^{\prime}$. Then
$\Lambda^{\prime}=\mathbb{Z}[\alpha^{\prime},\beta^{\prime},\gamma^{\prime\prime},\Lambda^{2}(\alpha^{\prime})]$
and, it is enough to check that
$\alpha^{\prime},\beta^{\prime},\gamma^{\prime\prime},\Lambda^{2}(\alpha^{\prime})$
are algebraically independent.

Using Proposition 4.2, we have
$\alpha^{\prime}=f(\lambda_{1})+\Delta_{5}^{-},\leavevmode\nobreak\
\beta^{\prime}=f(\lambda_{1})+\Delta_{5}^{+},\leavevmode\nobreak\
\gamma^{\prime\prime}=g(\lambda_{1})+\lambda_{2}$ and
$\Lambda^{2}(\alpha^{\prime})=h(\lambda_{1},\lambda_{2},\Delta_{5}^{-})+\lambda_{3}$,
where $f,g,h$ are polynomials in indicated variables. Suppose that
$F(\alpha^{\prime},\beta^{\prime},\gamma^{\prime\prime},\Lambda^{2}(\alpha^{\prime}))=0$.
Since $\lambda_{3},\lambda_{1},\lambda_{2},\Delta_{5}^{-},\Delta_{5}^{+}$ are
algebraically independent, and since $\Lambda^{2}(\alpha^{\prime})$ is
involves $\lambda_{3}$ and the others,
$\alpha^{\prime},\beta^{\prime},\gamma^{\prime\prime}$ do not involve
$\lambda_{3}$, we see that $F$ must be independent of
$\Lambda^{2}(\alpha^{\prime})$. Since there is no polynomial relation among
$\lambda_{2},\Delta_{5}^{+},\Delta_{5}^{-}$ over $\mathbb{Z}[\lambda_{1}]$, we
conclude that $F$ must be the zero polynomial. This proves the algebraic
independence of
$\alpha^{\prime},\beta^{\prime},\gamma^{\prime},\Lambda^{2}(\alpha^{\prime})$.

We have,
$\Delta_{5}^{-}=\alpha^{\prime}-f(\lambda_{1}),\Delta_{5}^{+}=\beta^{\prime}-f(\lambda_{1}),\lambda_{2}=\gamma^{\prime}-\alpha^{\prime}-\beta^{\prime}-g(\lambda_{1})$.
By substituting $\Delta_{5}^{-}$ and $\lambda_{2}$ from the above in
$\lambda_{3}=\Lambda^{2}(\alpha^{\prime})-\lambda_{1}-(1+\lambda_{1})\Delta_{5}^{-}-\lambda_{2}$
we see $R\mathrm{Spin}(10)\cong\Lambda^{\prime}[\lambda_{1}]$, as a
$\Lambda^{\prime}$-module. ∎

###  5.2. Computation of $\mathrm{Tor}^{*}_{R\mathrm{E}_{6}}(RH,\mathbb{Z})$

In order to compute the $E_{2}$-page of the Hodgkin spectral sequence
$\mathrm{Tor}^{*}_{R\mathrm{E}_{6}}(RH,\mathbb{Z})$, which converges to
$K^{*}(\mathrm{E}_{6}/\mathrm{Spin}(10))$, we will apply the change of ring
spectral sequence ([7, §5,Chapter XVI]), in the case following situation (in
the notation of [7]):

$\Gamma=R\mathrm{E}_{6},A=R\mathrm{Spin}(10),\leavevmode\nobreak\
K=C=\mathbb{Z}\textrm{\leavevmode\nobreak\ and\leavevmode\nobreak\
}\Lambda:=\mathbb{Z}[\alpha,\beta,\gamma,\Lambda^{2}(\alpha)]\subset\Gamma=R\mathrm{E}_{6}.$

Then $A$ is a free $\Lambda$-module via the restriction homomorphism, in view
of Lemma 5.2. Then

(15)
$\textrm{Tor}^{\Omega}_{*}(\textrm{Tor}_{*}^{\Lambda}(R\mathrm{Spin}(10),\mathbb{Z}),\mathbb{Z})\implies\textrm{Tor}^{R\mathrm{E}_{6}}_{*}(R\mathrm{Spin}(10),\mathbb{Z}),$

where
$\Omega=R\mathrm{E}_{6}/\langle\alpha-27,\beta-27,\gamma-78,\Lambda^{2}(\alpha)-351\rangle$.
In view of Lemma 5.2, we have
$\mathrm{Tor}^{\Lambda}_{*}(R\mathrm{Spin}(10),\mathbb{Z})=\mathrm{Tor}^{\Lambda}_{0}(R\mathrm{Spin}(10),\mathbb{Z})=R\mathrm{Spin}(10)\otimes_{\Lambda}\mathbb{Z}$
and the above spectral sequence collapses. So we have
$\mathrm{Tor}^{R\mathrm{E}_{6}}_{*}(R\mathrm{Spin}(10),\mathbb{Z})\cong\mathrm{Tor}^{\Omega}_{*}(B,\mathbb{Z})$
where

(16)
$B_{q}=\textrm{Tor}^{\Lambda}_{q}(R\mathrm{Spin}(10),\mathbb{Z})=\begin{cases}RH\otimes_{\Lambda}\mathbb{Z},&\textrm{if\leavevmode\nobreak\
}q=0,\\\ 0,\textrm{\leavevmode\nobreak\ if\leavevmode\nobreak\ }q\neq
0.\end{cases}.$

Thus,

$B=B_{0}=RH/\langle\alpha^{\prime}-27,\beta^{\prime}-27,\gamma^{\prime}-78,\Lambda^{2}(\alpha^{\prime})-{27\choose
2}\rangle.$

In view of Lemma 5.2, $B$ is isomorphic to the polynomial algebra
$\mathbb{Z}[\lambda_{1}].$

We shall continue to denote by $\lambda_{j},j=1,2,3$, the images in $B$ of the
elements $\lambda_{j}\in RH.$

######  Lemma 5.3.

The following relations hold in $B$:
(a) $\Delta_{5}^{+}=26-\lambda_{1}=\Delta_{5}^{-}$,
(b) $\lambda_{2}=25+2\lambda_{1}$,
(c) $\lambda_{3}=\lambda_{1}^{2}-28\lambda_{1}+300$,
(d) $\Lambda^{2}(\beta^{\prime})-{27\choose 2}=0$,
(e) $\Lambda^{2}(\gamma^{\prime})-{78\choose 2}=(\lambda_{1}-10)^{3}$.

###### Proof.

The proof of (a) and (b) are immediate from Proposition 4.2(i) and (ii),
respectively.

In view of Proposition 4.2 (iv), we have the relation ${27\choose
2}=\Lambda_{2}(\alpha^{\prime})=\lambda_{2}+\lambda_{3}+\lambda_{1}+(\lambda_{1}+1)\Delta_{5}^{-}$
in $B$. Using (a) and (b) we obtain ${27\choose
2}-\lambda_{3}=(25+2\lambda_{1})+(\lambda_{1}+1)(26-\lambda_{1})+\lambda_{1}.$
Now (c) follows up on simplification.

(d). Using Proposition 4.2(i) and (ii) again, we have
$\Delta_{5}^{+}-\Delta_{5}^{-}=\beta^{\prime}-\alpha^{\prime}=0$ in $B$. Now
Proposition 4.2 (iv) and (v) shows that
$\Lambda^{2}(\beta^{\prime})=\Lambda^{2}(\alpha^{\prime}).$ Therefore,
$\Lambda^{2}(\beta^{\prime})-{27\choose
2}=\Lambda^{2}(\alpha^{\prime})-{27\choose 2}=0$ in $B$.

(e) Recall from Proposition 4.2(iii), (vi) that the relations
$\gamma^{\prime}=1+\lambda_{2}+\Delta_{5}^{+}+\Delta_{5}^{-}$ and
$\Lambda^{2}(\gamma^{\prime})=1+2\lambda_{2}+(1+\lambda_{2})(\Delta_{5}^{+}+\Delta_{5}^{-})+2\lambda_{3}+\lambda_{1}\lambda_{3}$
hold in $\mathrm{Spin}(10).$ Using relations $\gamma^{\prime}=78$ and the
relations (a) to (d) of the lemma which have just been proved, we obtain

$\displaystyle\Lambda^{2}(\gamma^{\prime})$
$\displaystyle=(1+\lambda_{2}+\Delta_{5}^{+}+\Delta_{5}^{-})+\lambda_{2}(1+\Delta_{5}^{+}+\Delta_{5}^{-})+2\lambda_{3}+\lambda_{1}\lambda_{3}$
$\displaystyle=\gamma^{\prime}+\lambda_{2}(\gamma^{\prime}-\lambda_{2})+2\lambda_{3}+\lambda_{1}\lambda_{3}$
$\displaystyle=78+\lambda_{2}(78-\lambda_{2})+2\lambda_{3}+\lambda_{1}\lambda_{3}$
$\displaystyle=78+(2\lambda_{1}+25)(53-2\lambda_{1})+(\lambda_{1}+2)(\lambda_{1}^{2}-28\lambda_{1}+300)$
$\displaystyle=\lambda_{1}^{3}-30\lambda_{1}^{2}+300\lambda_{1}+2003$
$\displaystyle=(\lambda_{1}-10)^{3}+3003.$

Since $\gamma^{\prime}=78$ we have $\Lambda^{2}(\gamma^{\prime})={78\choose
2}=3003$ in $B$. The last equation now implies that $(\lambda_{1}-10)^{3}=0$.
∎

We are now ready to prove Theorem 1.1.

Proof of Theorem 1.1: Since $B_{q}=0$ for all $q>0$, the change of rings
spectral sequence
$\textrm{Tor}_{*}^{\Omega}(\textrm{Tor}^{\Lambda}(RH,\mathbb{Z}),\mathbb{Z})=\textrm{Tor}^{\Omega}_{*}(B_{*},\mathbb{Z})$
which converges to
$\textrm{Tor}^{R\mathrm{E}_{6}}_{*}(R\mathrm{Spin}(10),\mathbb{Z})$ collapses
and we get
$\mathrm{Tor}^{R\mathrm{E}_{6}}_{q}(R\mathrm{Spin}(10),\mathbb{Z})\cong\mathrm{Tor}^{\Omega}_{q}(B,\mathbb{Z})$,
where
$\Omega=R\mathrm{E}_{6}/\langle\alpha-27,\beta-27,\gamma-78,\Lambda^{2}(\alpha)-351\rangle=\mathbb{Z}[\Lambda^{2}(\beta),\Lambda^{2}(\gamma)]$.

Let us write $x=\Lambda^{2}(\beta)-{27\choose 2}$ and
$y=\Lambda^{2}(\gamma)-{78\choose 2}$. Then $\Omega=\mathbb{Z}[x,y]$, and the
Koszul resolution $\mathcal{C}$ of the polynomial ring $\Omega$ is

$0\to\Omega\xrightarrow{d}\Omega\oplus\Omega\xrightarrow{d}\Omega\xrightarrow{\epsilon}\mathbb{Z}\to
0$

where $\epsilon:\Omega\to\mathbb{Z}$ is the augmentation map defined as
$x\mapsto 0,y\mapsto 0$. We shall denote by $X,Y$ the basis of
$\Omega_{1}:=\Omega\oplus\Omega$ so that the basis of $\Omega_{2}\cong\Omega$
is $X\wedge Y.$ The differentials of the Koszul complex are defined by
$d(X)=x=\Lambda^{2}(\beta^{\prime})-{27\choose 2},\leavevmode\nobreak\
d(Y)=y=\Lambda^{2}(\gamma^{\prime})-{78\choose 2}$ and $d(X\wedge
Y)=d(X)Y-d(Y)X.$ Therefore, since $B$ is a $\Omega$-module via the
homomorphism $\bar{{\rho}}$ induced by $\rho$, on tensoring with $B$, we
obtain the following chain complex $\mathcal{C}\otimes B$:

$0\to B\xrightarrow{\bar{d}}B\oplus B\xrightarrow{\bar{d}}B\to 0$

where, in view of Lemma 5.3, $\bar{d}(X)=\bar{\rho}(x)=0,\leavevmode\nobreak\
\bar{d}(Y)=\bar{\rho}(y)=(\lambda_{1}-10)^{3}$ and $\bar{d}(X\wedge
Y)=\bar{\rho}(x)Y-X\bar{\rho}(y)=(\lambda_{1}-10)^{3}X$. Therefore,
$\mathrm{Tor}_{0}^{\Omega}(B,\mathbb{Z})=H^{0}(\mathcal{C}\otimes
B)=B/\langle(\lambda_{1}-10)^{3}\rangle$ and
$\mathrm{Tor}_{1}^{\Omega}(B,\mathbb{Z})=H^{1}(\mathcal{C}\otimes
B)=BX/\langle(\lambda_{1}-10)^{3}X\rangle\cong
B/\langle(\lambda_{1}-10)^{3}\rangle$. Let

(17) $u=\lambda_{1}-10.$

Since
$\mathrm{Tor}^{*}_{\Omega}(B,\mathbb{Z})=\mathrm{Tor}_{-*}^{\Omega}(B,\mathbb{Z})$
is generated, as a ring, by elements of degree $\geq-1,$ by [18, Proposition,
§1.3], we conclude that $\mathrm{Tor}_{*}^{\Omega}(B;\mathbb{Z})$ is
isomorphic to $K^{*}(\mathrm{E}_{6}/\mathrm{Spin}(10)).$ That is
$K^{0}(\mathrm{E}_{6}/\mathrm{Spin}(10))=\mathbb{Z}[u]/\langle u^{3}\rangle,$
and
$K^{1}(\mathrm{E}_{6}/\mathrm{Spin}(10))=K^{0}(\mathrm{E}_{6}/\mathrm{Spin}(10))\cdot
X$, the free $K^{0}(\mathrm{E}_{6}/\mathrm{Spin}(10))$ of rank $1$ generated
by $X.$ This completes the proof of Theorem 1.1. $\blacksquare$

###  5.3. Complexification and realification

Let $c:RO(G)\to RG$ and $r:RG\to RO(G)$ denote the complexification and the
‘forgetful’ homomorphism obtained by restricting the scalar multiplication
from $\mathbb{C}$ to $\mathbb{R}$ respectively. The homomorphism $r$ will be
referred to as the realification homomorphism. We shall use the same symbols
for the analogous homomorphisms $KO^{0}(X)\to K^{0}(X)$ and $K^{0}(X)\to
KO^{0}(X)$ for topological spaces. We note that in both contexts, $c$ is a
ring homomorphism, whereas $r$ is only a group homomorphism. As is well-known
$c\circ r=1+\bar{\leavevmode\nobreak\ }$, where in both contexts,
$\bar{\leavevmode\nobreak\ }$ denotes the complex conjugation, and $r\circ
c=\times 2$.

When $H\subset G$ is a closed subgroup of a compact connected Lie group and
$X=G/H,$ the maps $c$ and $r$ commute with the $\alpha$-construction, i.e.,
the following diagram commutes:

${RO(H)}$${RH}$${RO(H)}$${KO^{0}(G/H)}$${K(G/H)}$${KO^{0}(G/H)}$$\scriptstyle{\alpha_{0}}$$\scriptstyle{\alpha}$$\scriptstyle{\alpha_{0}}$
Figure 4.

For a more detailed discussion, see [14, §11, Chapter 13]. We remark that if
$[V]\in RO(H)$ is the restriction to $H$ of a $G$-representation, then
$\alpha_{0}([V])=\dim V$ [9, p.209, Example 2]. An analogous statement holds
for $\alpha.$

Since $G$ is compact, $RO(G)$ (resp. $RG$) is a free abelian group with basis
the set of isomorphism classes of irreducible $G$-representation over
$\mathbb{R}$ (resp. $\mathbb{C}$), and since $c\circ r=\times 2$, $c$ is a
monomorphism.

Also,
$c(\Lambda_{\mathbb{R}}^{j}([V]))=\Lambda^{j}_{\mathbb{C}}(c[V])\leavevmode\nobreak\
\forall j\geq 0.$ for any real representation $V$. Thus $c$ is a
$\lambda$-ring homomorphism. (However,
$\mathbb{R}=\Lambda^{2}(r(\mathbb{C}))\neq
r(\Lambda^{2}(\mathbb{C}))=r(0)=0$.) Analogous statements hold for vector
bundles. These properties will be used below without explicit reference.

###  5.4. The tangent bundle of $\mathrm{E}_{6}/\mathrm{Spin}(10)$

Let us denote by $\lambda_{1,\mathbb{R}}\in RO(\mathrm{Spin}(10))$ the class
of the standard representation of $\mathrm{SO}(10)$ on $\mathbb{R}^{10}$
regarded as a representation of $\mathrm{Spin}(10)$. Denote by
$\lambda_{j,\mathbb{R}}$ the representation
$\Lambda^{j}(\lambda_{1,\mathbb{R}})$. Thus
$\lambda_{j,\mathbb{R}}\otimes\mathbb{C}=\lambda_{j}\in R(\mathrm{Spin}(10)).$
Note that $\lambda_{2,\mathbb{R}}=Lie(\mathrm{Spin}(10))$, the adjoint
representation of $\mathrm{Spin}(10).$

######  Proposition 5.4.

Let $\tau$ be the tangent bundle of $\mathrm{E}_{6}/\mathrm{Spin}(10)$. Then

$[\tau]=[53-2\lambda_{1,\mathbb{R}}]\in KO(\mathrm{E}_{6}/\mathrm{Spin}(10)).$

###### Proof.

For any homogeneous space $M=G/H$ one has the isomorphism $\tau$ is isomorphic
to the vector bundle associated to the representation
$\alpha_{0}(Lie(G)/Lie(H))$ of $H$. Therefore, we have (see §5.3)

(18)
$[\tau]=\alpha_{0}[Lie(E_{6})]-\alpha_{0}[Lie(\mathrm{Spin}(10)]=78-\alpha_{0}[Lie(\mathrm{Spin}(10))]=78-\lambda_{2,\mathbb{R}}.$

By Proposition 4.2 (iii),
$Lie(\mathrm{E}_{6})\otimes\mathbb{C}=V_{\varpi_{5}}=\gamma\in
R\mathrm{E}_{6}$ and restricts to
$1+\lambda_{2}+\Delta_{5}^{+}+\Delta_{5}^{-}$ as a representation of
$\mathrm{Spin}(10)$. Now
$1+\lambda_{2}+\Delta^{+}_{5}+\Delta^{-}_{5}=c(1+\lambda_{2,\mathbb{R}}+r(\Delta^{+}_{5}))$
since $\Delta^{-}_{5}$ is the complex conjugate of $\Delta^{+}_{5}$. So
$c([Lie(\mathrm{E}_{6})-Lie(\mathrm{Spin}(10))])=c(1+r(\Delta^{+}_{5}))$ in
$R\mathrm{Spin}(10).$ Since $c:RO(\mathrm{Spin}(10))\to R\mathrm{Spin}(10)$ is
a monomorphism, we have
$[Lie(\mathrm{E}_{6})-Lie(\mathrm{Spin}(10))]=1+r(\Delta^{+}_{5})\in
RO(\mathrm{Spin}(10)).$ Set $[V]:=r(\Delta^{+}_{5})\in RO(\mathrm{Spin}(10))$.
Then

(19) $[\tau]=[1+V]\in KO(\mathrm{E}_{6}/\mathrm{Spin}(10)).$

Recall that $\rho(V_{\varpi_{1}})=1+\lambda_{1}+\Delta^{+}_{5}$ in
$R\mathrm{Spin}(10)$ by Proposition 4.2(i), and so
$r(V_{\varpi_{1}})=2+2\lambda_{1,\mathbb{R}}+V$. Since $r(\alpha^{\prime})$ is
restriction of the representation $r(\alpha)$ of $\mathrm{E}_{6}$ to
$\mathrm{Spin}(10)$, it follows that the class of the real vector bundle
associated to $2+2\lambda_{1}+V$ in $KO(\mathrm{E}_{6}/\mathrm{Spin}(10))$ is
trivial of rank $54$, that is, $[2+2\lambda_{1,\mathbb{R}}+V]=[54]\in
KO(\mathrm{E}_{6}/\mathrm{Spin}(10))$. Substituting for $[V]$ in the
expression for $[\tau]$ in Equation (19), we obtain that
$[\tau]+2[\lambda_{1,\mathbb{R}}]=53$. ∎

We are ready to prove Theorem 1.2.

###### Proof.

By Proposition 5.4, we have $[\tau]+[2\lambda_{1,\mathbb{R}}]=53.$ Since
$\mathrm{E}_{6}/\mathrm{Spin}(10)$ has dimension $33$, we obtain an
isomorphism of vector bundles $\tau\oplus\nu\cong 53\epsilon$ where $\nu$ is
associated to $2\lambda_{1,\mathbb{R}}.$ (Cf. [14, Chapter 9, Theorem 1.5].)
It follows from Hirsch’s immersion theorem [12], [14, Chapter 18, Remark 2.7],
that $\mathrm{E}_{6}/\mathrm{Spin}(10)$ can be immersed in $\mathbb{R}^{53}.$

Singhof and Wemmer ([20, Theorem 2]) have shown that
$M=\mathrm{E}_{6}/\mathrm{Spin}(10)$ is not stably parallelizable. In fact,
they show (ibid, §3) that the rational Pontjagin class $p_{2}(M)$ does not
vanish. Also Singhof ([19, Cor. 3.3(a), p.105]) has shown that $p_{1}(M)=0$ It
follows that $p_{2}(\nu)=-p_{2}(\tau M)\neq 0$. This yields a lower bound for
the immersion dimension of $M$, namely $M$ that cannot be immersed in
$\mathbb{R}^{40}.$ ∎

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